Motion

"Motion is change of position of an object with time."

1.0Motion

If we look around us, we find that there are number of objects which are in motion. An object is said to be in motion if it changes its position with the passage of time. Now observe the following bodies or objects to understand the meaning of the term "motion". Cars, cycles, motorcycles, scooters, buses, rickshaws, trucks, etc. running on the road, birds flying in the sky, fish swimming in water, all these objects are in motion. Very small objects like atoms, molecules and very large objects like planets, stars and galaxies are also in motion. Thus, all objects ranging from the smallest atom to the largest galaxy are in continuous motion. Kinematics is the science of describing the motion of objects using words, diagrams, numbers, graphs and equations. "Motion is the change in position of an object with time." Concept of a point object (or particle)

Point object

An extended object can be treated as a point object when the distance travelled by the object is much greater than its own size. A point object (or particle) is one which has no linear dimensions but possesses mass. Examples: (i) Study of motion of a train travelling from Kota to New Delhi. (ii) Revolution of earth around the sun for one complete revolution.

Describing motion

When a tree is observed by an observer A standing at the railway station, the tree is at rest. This is because position of the tree is not changing with respect to the observer A (see figure).

Now, when the same tree is observed by an observer B sitting in a superfast train moving with a velocity v , then the tree is moving with respect to the observer because the position of tree is changing with respect to the observer B. Rest and motion are relative terms : There is nothing like absolute rest or motion. This means that an object can be at rest and also in motion at the same time i.e. all objects, which are stationary on earth, are said to be at rest with respect to each other, but with respect to the sun they are making revolutions. In order to study motion, therefore, we have to choose a fixed position or point with respect to which the motion has to be studied. Such a point or fixed position is called a reference point or the origin.

Discuss whether the walls of your classroom are at rest or in motion.

The walls of our classroom are at rest with respect to the ground or earth. But, they are in motion with respect to an object or an observer outside the earth. This is because the earth is moving about its own axis as well as it is revolving around the sun. Thus, the state of rest and motion are not absolute, they are relative terms.

2.0Scalar and vector quantities

Scalar quantity

A physical quantity that is defined by its magnitude only is called a scalar quantity. Examples : Mass, time, distance, speed, work, power, energy, electric charge, volume, density, pressure, electric potential, temperature, etc.

Scalar quantities follow the algebraic (scalar) laws of addition.

Vector quantity

A physical quantity that is defined by its magnitude as well as direction is called a vector quantity. Examples : Velocity, acceleration, force, displacement, momentum, weight, torque, electric field, magnetic field, etc. Vector quantities follow the vector laws of addition. Arrows (or rays) are used to represent vectors. The direction of the arrow gives the direction of the vector. The length of the arrow is proportional to the magnitude of the vector.

Difference between scalar & vector quantities

3.0Distance and displacement

Distance

The length of the actual path between the initial and the final position of a moving object in the given time interval is known as the distance travelled by the object. Distance $=$ Length of path I (ACB) (see figure) $\sigma$ Distance is a scalar quantity. It is always taken positive. Distance is measured by odometer in vehicles. Units In SI system : metre (m). In CGS system : centimetre (cm).

Displacement

The shortest distance between the initial position and the final position of a moving object in the given interval of time is known as the displacement of the object. Displacement = Length of path II (AB) (see figure) Displacement of an object may also be defined as the change in position of the object in a particular direction. That is, Displacement of an object $=$ Final position - Initial position of the object $={\mathbf{x}}_{\mathbf{f}}-{\mathbf{x}}_{\mathbf{i}}$

• Displacement is a vector quantity. Displacement can be positive, negative or zero. Units In SI system : metre (m) In CGS system : centimetre (cm) Important points related to distance and displacement During motion, displacement of an object may be zero but the distance travelled by the object is never zero.
  
  Sign convention for displacement

Distance travelled by an object is either equal to or greater than the magnitude of displacement of the object. (G) Distance is equal to magnitude of displacement when a body moves in a straight line in a particular direction or it is in uniform motion.

A honeybee leaves the hive and travels 2 m as it returns to the hive. Is the displacement for the trip the same as the distance travelled? If not, why not?

Hive

Numerical Ability

• Motion of a particle is shown below on a number line. Find the displacement from (a) A to B (b) B to C (c) overall journey. Also, find distance for overall journey.
  
  Decode the problem: To find the displacement, we should know the initial and the final position of the particle. Apply the formula - Displacement $={\mathbf{x}}_{\mathbf{f}}-{\mathbf{x}}_{\mathbf{i}}$ Distance $=$ Length of actual path travelled Solution: (a) At point A, initial position of the particle ( ${\mathrm{x}}_{\mathrm{i}}$ ) is 0 m and at point $B$, the final position of the particle $(xf)$ is +6 m . Displacement from $A$ to $B=xf-x_i=(6)-(0)=+6m$ (b) At point B, initial position of the particle ( ${\mathrm{x}}_{\mathrm{i}}$ ) is 6 m and at point $C$, the final position of the particle $\left({\mathrm{x}}_{\mathrm{f}}\right)$ is -6 m . Displacement from B to $C=x_f-x_i=(-6)-(6)=-12\mathbf{m}$ (c) At point A, initial position of the particle ( ${\mathrm{xi}}_{\mathrm{i}}$ ) is 0 m and at point C , the final position of the particle $\left({\mathrm{x}}_{\mathrm{f}}\right)$ is -6 m . Displacement of overall journey (i.e. A to B, B to C) $=x_f-x_i=(-6)-(0)=-6\mathbf{m}$ Here, distance can also be found by adding positive values of displacement AB & displacement BC . i.e., Distance travelled during overall journey $=AB+BC$ $=6+12=18\mathrm{m}$ Here Distance > |Displacement|
• A body falls from a height of (\mathbf{3\mathbf{~ m}). Find displacement and distance.
  
  Solution- The initial position of the particle ( ${\mathrm{xi}}_{\mathrm{i}}$ ) is 3 m (the body is 3 m above the ground) and the final position of the particle ( xf ) is 0 m . (see figure) Displacement $=x_f-x_i=0-3=-3m$, Distance $=3\mathrm{m}$ (see figure)
• A particle moves along a circular path as shown in figure. Find distance travelled and displacement.
  
  Decode the problem: To find the displacement, we should know the initial and the final position of the particle. Apply the formula - Displacement $={\mathbf{x}}_{\mathbf{f}}-{\mathbf{x}}_{\mathbf{i}}$ Distance $=$ Length of actual path travelled Solution- The shortest distance is the straight line between points $A$ and $B$. Displacement $=$ diameter $\mathrm{AB}=2R$ To find the distance, we should know the formula of the circumference of the circle. Circumference of the circle $=2\pi r$ Distance travelled $=\frac{1}{2}\times$ (circumference of the circle $)$ $=\frac{1}{2}(2\pi \mathrm{R})=\pi \mathrm{R}$

Remember

• Always take care of the sign convention while finding the displacement of the body.
• Always take care of the direction (north, south, east, west) given in the problem.
  
• For a Circle : Diameter $=2\mathrm{r}$ Circumference $=2\pi$ r or $\pi \mathrm{d}(\pi =\frac{22}{7}$ or 3.14$)$
  
• Quadrilateral Perimeter of quadrilateral $=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}$ Perimeter of rectangle $=2(l+b)$ Perimeter of square $=4\times$ side
  
• Pythagoras theorem $(\text{ Hypotenuse }{)}^2=(\text{ Perpendicular }{)}^2+(\text{ Base }{)}^2$ ${\mathrm{AC}}^2={\mathrm{BC}}^2+{\mathrm{AB}}^2$
  
• Whenever a particle changes its direction, distance is greater than displacement.
• A body moving in a circular path when reaches its original position after one round, then the displacement at the end of one round is zero, but the distance travelled by it is equal to the circumference of circular path.
• A body moving in a circular path when covers $\frac{1}{4}$ th of the circle then, the distance travelled by it is $\frac{\pi r}{2}$ and displacement is $\sqrt{2\mathrm{r}}$.
• A body moving in a circular path when covers $\frac{3}{4}$ th of the circle then, the distance travelled by it is $\frac{3\pi \mathrm{r}}{2}$ and displacement is $\sqrt{2\mathrm{r}}$.

Comparison between distance and displacement

4.0Uniform and non-uniform motion

A moving body may cover equal distances in equal intervals of time or different distances in equal intervals of time. On the basis of above assumption, the motion of a body can be classified as uniform motion and non-uniform motion.

Uniform motion

When a body covers equal distances in equal intervals of time, however small may be the time intervals, in a particular direction, the body is said to describe a uniform motion. (see figure).

• Uniform motion always takes place in a straight line. In uniform motion, velocity of particle remains constant i.e., its magnitude as well as direction are constant. In uniform motion, average speed/velocity is equal to instantaneous speed/velocity at any point of time.
• Examples of uniform motion

(i) An aeroplane flying at a speed of $600\mathrm{km}/\mathrm{h}$ along north.

(ii) A train running at a speed of $120\mathrm{km}/\mathrm{h}$ along east.

(iii) Light energy travelling at a speed of $3\times 10^8\mathrm{m}/\mathrm{s}$ in vacuum.

Non-uniform motion

When a body covers unequal distances in equal intervals of time, the body is said to be moving with a non-uniform motion. (see figure) © Any motion along a curved path is always non-uniform motion. Also, any motion in which particle changes its direction is also non-uniform motion.

Distance-time graph for non-uniform motion

• Examples of non-uniform motion

(i) An aeroplane running on a runway before taking off. (ii) A freely falling stone under the action of gravity. (iii) When the brakes are applied to a moving car. (iv) A fan rotating with constant speed is also a non-uniform motion.

5.0Speed

Speed of a body is the distance travelled by the body per unit time. The rate of change of distance is called speed. $\text{ speed= }=\frac{\text{ distance travelled }}{\text{ time taken }}$

If a body covers a distance $s$ in time $t$ then speed, $\mathrm{v}=\frac{\mathrm{s}}{\mathrm{t}}$

Unit

In SI system : m/s or ${\mathbf{ms}}^{-1}$; In CGS system : cm/s or ${\mathbf{cms}}^{-1}$ A commonly used unit of speed is $\mathrm{km}/\mathrm{h}$ or ${\mathrm{kmh}}^{-1}$. © Speed is a scalar quantity because it has magnitude but no direction. Speed is always taken positive.

Uniform speed

When a body covers equal distances in equal intervals of time, the body is to be moving with a uniform speed or constant speed. Examples: (i) A train running with a speed of $120\mathrm{km}/\mathrm{h}$. (ii) An aeroplane flying with a speed of $600\mathrm{km}/\mathrm{h}$.

Non-uniform speed

When a body covers unequal distances in equal intervals of time, the body is said to be moving with non-uniform speed or variable speed. Examples : (i) A car running on a busy road. (ii) An aeroplane landing on a runway.

Average speed

The average speed of the body in a given time interval is defined as the total distance travelled, divided by total time taken. Average speed $=\frac{\text{ Total distance travelled }}{\text{ Total time taken }}$ $\frac{1\mathrm{km}}{\mathrm{h}}=\frac{1000\mathrm{m}}{60\times 60\mathrm{s}}=\frac{5}{18}\mathrm{m}/\mathrm{s}$ Speed in $\mathrm{m}/\mathrm{s}=\frac{5}{18}\times$ speed in $\mathrm{km}/\mathrm{h};\mathrm{Km}{\mathrm{h}}^{-1}\underset{\times \frac{\times \frac{5}{5}}{18}}{\rightleftarrows }{\mathrm{ms}}^{-1}$

Numerical Ability

• A car travels first half distance with a uniform speed $u$ and next half distance with a uniform speed v. Find its average speed.

Solution:

Total distance $=\frac{\mathrm{d}}{2}+\frac{\mathrm{d}}{2}=\mathrm{d}$ [See figure] Total time $={\mathrm{t}}_1+{\mathrm{t}}_2=\mathrm{t}$ $\therefore {\mathrm{t}}_1=\frac{\mathrm{d}/2}{\mathrm{u}}$ $t_2=\frac{d/2}{v}$ $V_{av}=\frac{d}{t}=\frac{d}{t_1+t_2}$ Putting the value of equation (i) and (ii), ${\mathrm{V}}_{\mathrm{av}}=\frac{\mathrm{d}}{\frac{\mathrm{d}/2}{\mathrm{u}}+\frac{\mathrm{d}/2}{\mathrm{v}}}=\frac{\mathrm{d}}{\frac{\mathrm{d}}{2}\left(\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\right)}$ (Taking $\mathrm{d}/2$ common out of bracket in the denominator) $=\frac{2}{\frac{v+u}{uv}}$ $\therefore {\mathrm{V}}_{\mathrm{av}}=\frac{2\mathrm{uv}}{\mathrm{u}+\mathrm{v}}$

• A car travels first half time with a uniform speed $u$ and next half time with a uniform speed v. Find its average speed.

Decode the problem: Identify the type of motion Uniform motion or Non-uniform motion Identify the formula Average speed $=$ Total distance travelled Total time taken Total time $=\mathrm{t}$ Average speed $=\frac{\text{ Total distance }}{\text{ Total time taken }}$ Solution: Since the car travels with a constant speed therefore, to find the distance the formula to be used is Distance $=$ speed $\times$ time. $d_1=u\times \frac{t}{2},d_2=v\times \frac{t}{2}[\text{ See figure }]$

Instantaneous speed

The speed of a body at any particular instant of time during its motion is called the instantaneous speed of the body. It is measured by speedometer in vehicles.

• If body covers equal distances ${\mathrm{x}}_1={\mathrm{x}}_2=\mathrm{x}$ (let), with different speeds i.e., u and v , then directly use the given formula, ${\mathrm{v}}_{\text{average }}=\frac{2uv}{u+v}$
• If the two time intervals are same i.e., ${\mathrm{t}}_1={\mathrm{t}}_2=\mathrm{t}$ (let), speeds are different then directly use the given formula, ${\mathrm{v}}_{\text{average }}=\frac{\mathrm{v}+\mathrm{u}}{2}$
• If the total distance or displacement is given then use the given formula, ${\mathrm{v}}_{\text{avg speed }}=\frac{\text{ Total distance }}{\text{ Total time }}{\mathrm{v}}_{\text{avg velocity }}=\frac{\text{ Total displacement }}{\text{ Total time }}$

Numerical Ability

• The average speed of a bicycle, an athlete and a car are $18\mathrm{km}/\mathrm{h},7\mathrm{m}/\mathrm{s}$ and $2\mathrm{km}/\min$. respectively. Which of the three is the fastest and which is the slowest?

Decode the problem To find the fastest and slowest speed, we need to convert all the given speeds in SI units. Identify the formula $\mathrm{Km}{\mathrm{h}}^{-1}\underset{\times \frac{\times \frac{5}{5}}{\stackrel{\times 18}{\rightleftarrows }}}{\stackrel{\circ }{\leftrightarrows }}{\mathrm{ms}}^{-1}$ Solution: $18\mathrm{km}/\mathrm{h}=\frac{18\mathrm{km}}{1\mathrm{h}}=\frac{18000\mathrm{m}}{3600\mathrm{s}}=5\mathrm{m}/\mathrm{s}$ $2\mathrm{km}/\min =\frac{2\mathrm{km}}{1\min .}=\frac{2000\mathrm{m}}{60\mathrm{s}}=33.3\mathrm{m}/\mathrm{s}$ Thus, the average speeds of the bicycle, the athlete and the car are $5\mathrm{m}/\mathrm{s},7\mathrm{m}/\mathrm{s}$ and $33.3\mathrm{m}/\mathrm{s}$ respectively. So the car is the fastest, and the bicycle is the slowest.

• On a 120 km track, a train travels the first 30 km with a uniform speed of $30\mathbf{km}/\mathrm{h}$. How fast must the train travel the next 90 km so as to average $60\mathrm{km}/\mathrm{h}$ for the entire trip? Solution: Decode the problem Identify the type of motion Uniform motion or Non-uniform motion Identify the formula Average speed $=$ Total distance travelled Total time taken Total time $=\mathrm{t}$ Average speed $=\frac{\text{ Total distance }}{\text{ Total time taken }}$ Given: Total distance $\mathrm{d}=120\mathrm{km}$, Average speed ${\mathrm{V}}_{\mathrm{av}}=60$ $\mathrm{km}/\mathrm{h}$, Total time $=\mathrm{t}=$ ? ${\mathrm{V}}_{\mathrm{av}}=\frac{\mathrm{d}}{\mathrm{t}}$ or $\mathrm{t}=\frac{\mathrm{d}}{{\mathrm{V}}_{\mathrm{av}}}$ Putting the values, $\mathrm{t}=\frac{120\mathrm{km}}{60\mathrm{km}/\mathrm{h}}=2\mathrm{h}$ Distance travelled in first part of trip, ${\mathrm{d}}_1=30\mathrm{km}$, Speed in first part of the trip, ${\mathrm{v}}_1=30\mathrm{km}/\mathrm{h}$ Time taken in first part of trip, ${\mathrm{t}}_1=$ ?, ${\mathrm{t}}_1=\frac{{\mathrm{d}}_1}{{\mathrm{v}}_1}$ Putting the values, $t_1=\frac{30\mathrm{km}}{30\mathrm{km}/\mathrm{h}}=1\mathrm{h}$ Time taken to complete second part of the trip, ${\mathrm{t}}_2=\mathrm{t}-{\mathrm{t}}_1=2-1=1\mathrm{h}$ Distance to be covered in second part of the trip, ${\mathrm{d}}_2=90\mathrm{km}$, Required speed in second part, ${\mathrm{v}}_2=$ ? Speed $=\frac{\text{ distance }}{\text{ time }}$, $\therefore \mathrm{v}2=\frac{{\mathrm{d}}_2}{{\mathrm{t}}_2}=\frac{90\mathrm{km}}{1\mathrm{h}}=90\mathrm{km}/\mathrm{h}$ Check the unit Use $\mathrm{km}/\mathrm{hr}$ as the distance is given in km and time is given in hrs.
• A bus going from Kota to Jaipur passed the $100\mathrm{km},160\mathbf{km}$ and 220 km points at 10:30 am, 11:30 am and 1:30 pm. Find the average speed of the bus during each of the following intervals: (a) 10.30 am to 11.30 am , (b) 11.30 am to 1.30 pm and (c) 10.30 am to 1.30 pm. Solution: (a) The distance covered between 10:30 am and 11:30 am is $160\mathrm{km}-100\mathrm{km}=60\mathrm{km}$. The time interval is 1 hour. The average speed during this interval is - ${\mathrm{v}}_1=\frac{60\mathrm{km}}{1\mathrm{h}}=60\mathrm{km}/\mathrm{h}$

Decode the problem Identify the type of motion Uniform motion $✓/$ Non-uniform motion Identify the formula Average speed $=$ Total distance travelled Total time taken (b) The distance covered between 11:30 am and 1:30 pm is $220\mathrm{km}-160\mathrm{km}.=60\mathrm{km}$. The time interval is 2 hours. The average speed during this interval is - ${\mathrm{v}}_2=\frac{60\mathrm{km}}{2\mathrm{h}}=30\mathrm{km}/\mathrm{h}$ (c) The distance covered between 10:30 am and 1:30 pm is $220\mathrm{km}-100\mathrm{km}=120\mathrm{km}$. The time interval is 3 hours. The average speed during this interval is - ${\mathrm{v}}_3=\frac{120\mathrm{km}}{3\mathrm{h}}=40\mathrm{km}/\mathrm{h}$

6.0Velocity

The velocity of a body is the displacement of a body per unit time. $\text{ Velocity }=\frac{\text{ Displacement }}{\text{ Time }}$

The displacement covered by a body per unit time or the speed of a body in specified direction is called velocity. $\varpi$ Velocity is a vector quantity. It can be positive, negative or zero (see figure).

Unit

In SI system: $m/s$ or $m{s}^1$

In CGS system: $cm/s$ or $cm{s}^{-1}$

Uniform velocity

When a body covers equal displacements in equal intervals of time in a particular direction, the body is said to be moving with a uniform velocity (see figure).

Body moving with uniform velocity

Conditions for uniform velocity

(i) The body must cover equal displacements in equal intervals of time. (ii) The direction of motion of the body should not change.

Example : A train running towards south with a speed of $120\mathrm{km}/\mathrm{h}$.

Non-uniform velocity/variable velocity

When a body covers unequal displacements in equal intervals of time, the body is said to be moving with variable velocity (see figure).

When a body covers equal distances in equal intervals of time, but its direction changes, then the body is said to be moving with variable velocity.

Example : In circular motion, a particle may have constant speed but its direction changes continuously thus, its velocity is non-uniform (see figure).

Body moving with variable velocity

Conditions for variable velocity

(i) It should cover unequal displacements in equal intervals of time. (ii) It should cover equal distances in equal intervals of time but its direction must change.

Examples: (i) A car running towards north on a busy road has a variable velocity as the displacement covered by it per unit time changes with change in the road condition. (ii) The blades of a rotating ceiling fan, a person running around a circular track with constant speed etc. are examples of variable velocity.

Average velocity

Total displacement of a particle divided by total time taken is called average velocity.

• Average speed is always greater than or equal to magnitude of average velocity. Average speed is equal to average velocity when particle moves in a straight line without change in direction.

Instantaneous velocity

The velocity of a body at any particular instant of time during its motion is called the instantaneous velocity of the body.

Numerical Ability

• A particle moves along a path ABC as shown in figure. The time taken during the journey is 2 seconds. Find the average speed and average velocity during the journey.
  
  Solution: Decode the problem Identify the type of motion Uniform motion / Non-uniform motion Identify the formula Average speed = Total distance travelled Total time taken Identify the formula for Average velocity =

Total displacement travelled Total time taken Identify the initial position and the final position and the shortest distance between them. $\infty$ Here, average speed is greater than average velocity because the direction of particle Total distance travelled, $\mathrm{s}=\mathrm{AB}+\mathrm{BC}=6+8=14\mathrm{m}$ Average speed, ${\mathrm{V}}_{\mathrm{av}}=\frac{\mathrm{s}}{\mathrm{t}}=\frac{14}{2}=7\mathrm{m}/\mathrm{s}$ To find the displacement, we should know the initial and the final position of the particle. In a right angle triangle, $=\sqrt{(6{)}^2+(8{)}^2}=\sqrt{100}=10\mathrm{m}$ $\therefore$ Average velocity, ${\overrightarrow{\mathrm{V}}}_{\mathrm{av}}=\frac{\overrightarrow{\mathrm{s}}}{\mathrm{t}}=\frac{10}{2}=5\mathrm{m}/\mathrm{s}$ changes during motion.

Comparison between speed and velocity

7.0Acceleration

In uniform motion, the velocity remains constant with time. Thus, the change in velocity for any time interval is zero. But, in non-uniform motion, velocity changes with time. Thus, the change in velocity for any time interval has a non zero value.

In non-uniform motion, a new physical quantity called 'acceleration' is used. The rate of change of velocity of a moving body with time is called acceleration. $\text{ Acceleration }=\frac{\text{ Change in velocity }}{\text{ Time taken for change }}$

But, change in velocity $=$ final velocity - initial velocity. $\text{ Acceleration }=\frac{\text{ Final velocity }-\text{ Initial velocity }}{\text{ Time taken for change }}=\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}$

If body moves with uniform velocity, then $\mathrm{v}=\mathrm{u}$ and then acceleration is zero i.e. $\mathrm{a}=0$. Acceleration is a vector quantity. It can be negative, positive or zero (see figure).

Unit of acceleration

In SI system : m/s ${\mathbf{s}}^2$ or ms ${\mathbf{m}}^2$, In CGS system : cm/s ${\mathbf{s}}^2$ or ${\mathbf{cms}}^{-2}$ If the velocity of an object increases with time, such a motion is called 'accelerated motion'. In such motion, acceleration 'a' is considered positive for numerical problems.

Example : An object starts from rest and its velocity goes on increasing with time.

If velocity of an object decreases with time, such a motion is called 'retarded motion'. In such a motion, acceleration is called 'retardation' or 'deceleration' and it is considered negative for numerical problems. Example : A vehicle in motion is stopped by applying brakes.

Uniform acceleration

When a body moving in a straight line undergoes equal changes of velocity in equal intervals of time, the body is said to be moving with a uniform acceleration. Also, uniform acceleration means an acceleration having a constant magnitude and a constant direction (see figure).

Uniformly accelerating body Examples: (i) Motion of a freely falling body. (ii) Motion of a ball rolling down on an inclined plane.

Non-uniform acceleration or variable acceleration

When a body undergoes unequal changes of velocity in equal intervals of time, the body is said to be moving with non-uniform acceleration (see figure).

Non-uniformly accelerating body Examples: (i) The motion of a bus leaving or entering the bus stop. (ii) A car moving on a busy road has non-uniform acceleration.

Numerical Ability

• A scooter acquires a speed of $36\mathrm{km}/\mathrm{h}$ in 10 s after the start. Calculate the acceleration of the scooter.

Solution: Decode the problem Identify the type of motion Uniform motion / Non-uniform motion Identify the formula $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}$ Given, Initial velocity, $u=0$. Final velocity, v $=36\mathrm{km}/\mathrm{h}$ $=\frac{36\times 1\mathrm{km}}{1\mathrm{h}}=\frac{36\times 5\mathrm{m}}{18\mathrm{s}}=10\mathrm{m}/\mathrm{s}$ Time taken, $\mathrm{t}=10\mathrm{s}$ $\mathrm{a}=\frac{10\mathrm{m}/\mathrm{s}-0\mathrm{m}/\mathrm{s}}{10\mathrm{s}}=\frac{10\mathrm{m}/\mathrm{s}}{10\mathrm{s}}=1\mathrm{m}/{\mathrm{s}}^2$ Acceleration $=1\mathbf{m}/{\mathbf{s}}^2$

• The velocity of a runner is $10\mathrm{m}/\mathrm{s}$. After 4 seconds, his velocity becomes $20\mathrm{m}/\mathrm{s}$. What is the value of acceleration? Solution: Decode the problem Identify the type of motion Uniform motion / Non-uniform motion Identify the formula $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}$ Given, initial velocity, $u=10\mathrm{m}/\mathrm{s}$, final velocity, $\mathrm{v}=20\mathrm{m}/\mathrm{s}$; time, $\mathrm{t}=4$ seconds $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{20\mathrm{m}/\mathrm{s}-10\mathrm{m}/\mathrm{s}}{4\mathrm{s}}$ $=\frac{10\mathrm{m}/\mathrm{s}}{4\mathrm{s}}$ $=2.5\mathrm{m}/{\mathrm{s}}^2$ Acceleration $=2.5\mathbf{m}/{\mathbf{s}}^2$

8.0Equations of uniformly accelerated motion

When an object moves with a uniform acceleration, its motion is called 'uniformly accelerated motion'.

In case of uniformly accelerated motion, the average velocity is given by $V_{av}=\frac{v+u}{2}$, here $v=$ final velocity, $u=$ initial velocity These equations give the relationship between initial velocity, final velocity, time taken, acceleration and distance travelled by the body.

First equation of motion

A body having an initial velocity ' $u$ ' acted upon by a uniform acceleration 'a' for time ' t ' such that final velocity of the body is ' $v$ '.

Acceleration $=\frac{\text{ Change in velocity }}{\text{ Time taken }}=\frac{\text{ Final velocity }-\text{ Initial velocity }}{\text{ Time taken }}$ $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}$ or $\mathrm{v}-\mathrm{u}=\mathrm{at}$, or $\mathbf{v}=\mathbf{u}+\mathbf{at}$

Second equation of motion

It gives the distance travelled by a body in time ' t '. A body having an initial velocity 'u' acted upon by a uniform acceleration 'a' for time ' t ' such that final velocity of the body is ' $v$ ' and the distance covered is ' $s$ '. ${\mathrm{V}}_{\mathrm{av}}=\frac{\mathrm{v}+\mathrm{u}}{2}$ Distance covered $=$ average velocity $\times$ time taken $\mathrm{s}=\left(\frac{\mathrm{v}+\mathrm{u}}{2}\right)\times \mathrm{t}$ but $\mathrm{v}=\mathrm{u}+$ at (from first equation of motion) Using (1) & (2), we get, $s=\left(\frac{u+at+u}{2}\right)t=\left(\frac{2u}{2}+\frac{at}{2}\right)t=\left(u+\frac{at}{2}\right)t$ or $s=ut+\frac{1}{2}{\mathbf{at}}^2$

Third equation of motion

A body having an initial velocity ' $u$ ' moving with a uniform acceleration 'a' for time ' t ' such that final velocity is ' $v$ ' and the distance covered is ' $s$ '. ${\mathrm{V}}_{\mathrm{av}}=\frac{\mathrm{v}+\mathrm{u}}{2}$ Distance covered $=$ average velocity $\times$ time taken $\mathrm{s}=\left(\frac{\mathrm{v}+\mathrm{u}}{2}\right)\times \mathrm{t}$ ...(1) Now,

$v=u+at$ or $v-u=$ at

or

$t=\left(\frac{v-u}{a}\right)$ ...(2)

from (1) & (2), we get

$s=\left(\frac{v+u}{2}\right)\left(\frac{v-u}{a}\right)=\frac{v^2-u^2}{2a}$ or $2as=v^2-u^2$

or

$v^2=u^2+2as$

• While deriving the second equation of motion we eliminate final velocity (v) and replace it with $(u+at)$ as the second equation of motion gives the relation between the distance/displacement, initial velocity, acceleration and time.
• While deriving third equation of motion, we eliminate time ( t ) and replace it with $\left(\frac{v-u}{a}\right)$ as the third equation of motion gives the relation between the distance/displacement, initial velocity, acceleration and final velocity.

9.0Numerical Ability

• A car accelerates uniformly from $18{\mathrm{kmh}}^{-1}$ to $36{\mathrm{kmh}}^{-1}$ in $5$ s. Calculate (i) acceleration, (ii) distance covered by the car in that time. Solution: Decode the problem Identify the type of motion Uniform motion / Nonuniform motion $✓$ Identify the equation $\mathrm{v}=\mathrm{u}+\mathrm{at}✓$ (for part (i), since, $u,v$ and $t$ are given) $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2✓$ (for part (ii), since, $u$, a and $t$ are given) ${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$ Given, $\mathrm{u}=18\mathrm{km}/\mathrm{hr}=5\mathrm{m}/\mathrm{s}$; $\mathrm{v}=36\mathrm{km}/\mathrm{h}=10\mathrm{m}/\sec ;\mathrm{t}=5\mathrm{s}$ (i) We use first equation of motion

(ii) We use second equation of motion

• A car initially at rest starts moving with a constant acceleration of $0.5\mathrm{m}{\mathrm{s}}^{-2}$ and travels a distance of $25\mathbf{m}$. Find its final velocity. Solution: Decode the problem Identify the equation $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $s=ut+\frac{1}{2}{a}^2{}^2$ ${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}✓$ (since, $u$, a and $s$ are given) Given, $\mathrm{u}=0,\mathrm{a}=0.5\mathrm{m}/{\mathrm{s}}^2,\mathrm{s}=25\mathrm{m}$ We use third equation of motion $v^2=u^2+2as$ $v^2=(0{)}^2+2\times 0.5\times 25=25$ $\mathrm{v}=\sqrt{25}=5\mathrm{m}/\mathrm{s}$
• A racing car has a uniform acceleration of $5\mathrm{m}/{\mathrm{s}}^2$. What distance will it cover in $20\mathrm{s}$ after starting from rest? Solution: The car starts from the position of rest. So, its initial velocity $(\mathrm{u})$ is zero. Then, $u=0\mathrm{m}/\mathrm{s},\mathrm{a}=5\mathrm{m}/{\mathrm{s}}^2,\mathrm{t}=20\mathrm{s}$, $\mathrm{s}=$ ? For uniformly accelerated motion, we have three equations from which we use one equation. Decode the problem Identify the type of motion Uniform motion / Non-uniform motion Identify the equation $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $s=ut+\frac{1}{2}{a}^2$ $\mathrm{s}=(0\mathrm{m}/\mathrm{s}\times 20\mathrm{s})+\frac{1}{2}\times \left(5\mathrm{m}/{\mathrm{s}}^2\right)\times (20\mathrm{s}{)}^2$ or $\mathrm{s}=0\mathrm{m}+\frac{1}{2}\times 5\times 20\times 20$ $=5\times 10\times 20=1000\mathbf{m}=1\mathbf{km}$ So, the car will cover a distance of 1000 m (or 1 km ) in 20 s .
• A body starts from rest and has a velocity of $3\mathrm{m}/\mathrm{s}$. If it has a uniform acceleration of $10\mathrm{m}/{\mathrm{s}}^2$. Find the distance travelled by in 3 seconds. Solution: Decode the problem Identify the equation $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ ${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$ (since, $\mathrm{u},\mathrm{a}$ and s are given) The body starts from the position of rest. So, its initial velocity $(\mathrm{u})$ is zero. Then, $\mathrm{u}=0\mathrm{m}/\mathrm{s},\mathrm{a}=10\mathrm{m}/{\mathrm{s}}^2,\mathrm{t}=3\mathrm{s},\mathrm{s}=$ ?

• If a body has a velocity of $50\mathrm{m}/\mathrm{s}$, and it takes 4 seconds to travel a distance of 240 m , then what is the acceleration of the body? Solution: Decode the problem Identify the type of motion Uniform motion / Non-uniform motion Identify the equation $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $\mathrm{s}=\mathrm{ut}+\frac{1}{2}\mathrm{at}{\mathrm{t}}^2$ $v^2=u^2+2as$ (since, $\mathrm{u},\mathrm{a}$ and s are given) Given, initial velocity, $u=50\mathrm{m}/\mathrm{s}$, time, $\mathrm{t}=4\mathrm{s}$, distance, $\mathrm{s}=240\mathrm{m}$, acceleration, $\mathrm{a}=$ ? $240=(50\times 4)+\left(\frac{1}{2}\times \mathrm{a}\times 16\right);$ $240=200+8a\Rightarrow 40=8a$ $\mathrm{a}=5\mathbf{m}/{\mathbf{s}}^2$
• The speed of a car is reduced from $54\mathrm{km}/\mathrm{hr}$ to $36\mathrm{km}/\mathrm{hr}$ in a certain time during which in travelled 125 m . Calculate the acceleration of the car. Solution:

Decode the problem Convert speed from $\mathrm{km}/\mathrm{hr}$ to $\mathrm{m}/\sec$. Identify the type of motion Uniform motion / Nonuniform motion Identify the equation $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$ $v^2=u^2+2as$ (since, $u$, a and $s$ are given) Given, Initial velocity, $\mathrm{u}=54\mathrm{km}/\mathrm{h}=\frac{54\times 1\mathrm{km}}{1\mathrm{h}}=\frac{54\times 5\mathrm{m}}{18\mathrm{s}}$ $=15\mathrm{m}/\mathrm{s}$ Final velocity, $\mathrm{v}=36\mathrm{km}/\mathrm{h}$ $=\frac{36\times 1\mathrm{km}}{1\mathrm{h}}=\frac{36\times 5\mathrm{m}}{18\mathrm{s}}=10\mathrm{m}/\mathrm{s}$ Distance, $\mathrm{s}=125\mathrm{m}$; acceleration $=$ ? $a=\frac{v^2-u^2}{2s}=\frac{(10{)}^2-(15{)}^2}{2\times 125}$ $=\frac{100-225}{250}=\frac{-125}{250}=-0.5$ acceleration $=-0.5\mathbf{m}/{\mathbf{s}}^2$

• A train starting from rest attains a speed of $90\mathrm{km}/\mathrm{hr}$. If the acceleration of the train is $0.5\mathrm{m}/{\mathrm{s}}^2$, how much distance will it travel? Solution:

Given, initial velocity, $\mathrm{u}=0$; acceleration, $\mathrm{a}=0.5\mathrm{m}/{\mathrm{s}}^2$

Decode the problem Identify the type of motion Uniform motion / Nonuniform motion $✓$ Identify the formula $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $s=ut+\frac{1}{2}{a}^2{t}^2$ $v^2=u^2+2$ as$Finalvelocity,$$\mathrm{v}=90\mathrm{km}/\mathrm{h}=\frac{90\times 1\mathrm{km}}{1\mathrm{h}}=\frac{90\times 5\mathrm{m}}{18\mathrm{s}}=25\mathrm{m}/\mathrm{s}$$Foruniformlyacceleratedmotion,wehavethreeequationsfromwhichweuseoneequation.(Since$u, v$and$a$aregiven)Weusethirdequationofmotion$${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$$(25\mathrm{m}/\mathrm{s}{)}^2=0+2\times 0.5\times \mathrm{s}\Rightarrow 625=0+\mathrm{s}$\mathrm{s}=\mathbf{6 2 5} \mathrm{m}$

• The driver of a train travelling at $40{\mathrm{ms}}^{-1}$ applies the brakes as a train enters a station. The train slows down at a rate of $2{\mathrm{ms}}^{-2}$. How long will the train travel before it stops? Solution: Decode the problem Identify the type of motion Uniform motion / Nonuniform motion Identify the formula $v=u+at$ $s=ut+\frac{1}{2}{a}^2$ $v^2=u^2+2as$ Given, $\mathrm{u}=40\mathrm{m}/\mathrm{s},\mathrm{v}=0,\mathrm{a}=-2\mathrm{m}/{\mathrm{s}}^2$ (negative sign is taken as the velocity of object decreases with time and hence it is a retarded motion) $(0{)}^2=(40\mathrm{m}/\mathrm{s}{)}^2+2\times \left(-2\mathrm{m}/{\mathrm{s}}^2\right)\times \mathrm{s}$ $4\mathrm{m}/{\mathrm{s}}^2\times \mathrm{s}=40\times 40{\mathrm{m}}^2/{\mathrm{s}}^2$ $\mathrm{s}=\frac{1600}{4}$ $\mathrm{s}=400\mathrm{m}$

• The second equation of motion always tells the relation between the distance/displacement and time. If in a given problem these quantities are mentioned, then apply second equation of motion.
• The third equation of motion always tells the relation between the distance/displacement and final velocity. If in a given problem these quantities are mentioned, then apply third equation of motion.
• Squares of few numbers are given below. Learning these will make the calculations quick.

• Displacement, velocity and acceleration are vector quantities. Always take care of the sign convention of these quantities.
• If the final velocity is greater than the initial velocity, then acceleration is taken positive and if the final velocity is less then the initial velocity, then acceleration is taken negative in the numerical.
• Always take care of the units while solving the numerical. All the units should be in the same system of units.

Example : If the initial velocity of a body is $20\mathrm{m}/\mathrm{s}$, acceleration is $1\mathrm{m}/{\mathrm{s}}^2$, time is 2 min , then find the final velocity. Here, the time is to be converted into seconds, so that all the quantities are in same system of units and the final answer is also in the correct units.

10.0Graphical representation of motion

Graph

A graph is a line, straight or curved, showing the relation between two variable quantities, of which one varies as a result of the change in the other. The quantity which changes independently is called independent variable and the one which changes as a result of the change in the other is called dependent variable.

Distance/displacement-time graph

This graph is plotted between the time taken and the distance covered. The time is taken along the x -axis and the distance covered is taken along the y -axis.

• The slope of the distance-time graph gives the speed of the body.
• The slope of the displacement-time graph gives the velocity of the body.

When the body is at rest

When position of the body does not change with time then it is said to be stationary. The distance-time graph of such a body is a straight line parallel to $x$-axis (see figure).

When the body is in uniform motion

In uniform motion, $\mathrm{v}=$ constant. Now, $\mathrm{s}=\mathrm{v}\times \mathrm{t}$ Since v is constant $\therefore \mathrm{s}\propto \mathrm{t}$, thus, the distance-time graph of such a body is a straight line, inclined to x -axis. Slope $=\frac{\text{ measure on }y-\text{ axis }}{\text{ measure on }x-\text{ axis }}=\tan \theta$ or Slope $=\frac{x_2-x_1}{t_2-t_1}=v$

Thus, slope of distance-time graph gives speed of the body (see figure).

• In a distance-time graph, more the slope of the graph of an object in motion, more will be its speed and vice-versa. Example : In given graph (see figure a) speed of particle A is greater than the speed of particle B because slope of graph of particle A is greater than slope of graph of particle B.
• In the given graph (see figure b), speed of particle $A$ is equal to speed of particle $B$ because graphs of both have same slope.
  
  (a) Two particles moving with different velocities

When the body is in non-uniform motion

In this case, distance-time graph is a curve. For example, in uniformly accelerated motion, $s=ut+\frac{1}{2}a{t}^2$. Since $s\propto t^2$, definitely the graph is not a straight line, it is a curve (see figure).

A distance-time graph can never be parallel to y-axis (representing distance) because this line has inclination of $90^{\circ }$, and slope $=\tan \theta =\tan 90^{\circ }=$ infinite, which means infinite speed; it is impossible.

Speed/velocity-time graph

The variation in velocity with time for an object moving in a straight line can be represented by a velocity-time graph. In this graph, time is represented along the $x$-axis and velocity is represented along the $y$-axis. The slope of the speed/velocity-time graph gives the acceleration of the body. Area enclosed under a speed-time graph or velocity-time graph gives the distance covered by the body.

When the body is in uniform motion

Here, the body moves with constant velocity. The speed or velocity of the body is uniform, hence the magnitude remains same. The graph is a straight line parallel to x -axis

• Speed or velocity-time graph line can never be parallel to $y$-axis (speed axis), because inclination becomes $90^{\circ }$, then $\tan 90^{\circ }$ is infinite; ie. infinite acceleration is impossible.

Distance from speed or velocity-time graph

As distance $=$ speed $\times$ time, hence the distance can be calculated from speed or velocitytime graph.

When speed or velocity is uniform (constant) Distance $=$ Area of rectangle $\mathrm{ABCD}=\mathrm{AB}\times \mathrm{AD}$ [see figure (a)]

When acceleration is uniform (constant)

Distance or displacement = Area of right-angled triangle OAB [see figure (b)] $=\frac{1}{2}\times$ base $\times$ height $=\frac{1}{2}\times \mathrm{OB}\times \mathrm{BA}$

Area under v-t graph gives distance travelled by a body

• The unit of distance (m) is used to find the area under the curve of speed/velocity time graph. Distance $=$ Area under the curve in the above graph $=$ length (quantity on $y$-axis) $\times$ breadth (quantity on x -axis).

Distance $=$ Velocity $\times$ time $=\frac{\mathrm{m}}{\varphi }\times /,\varphi =\mathrm{m}$

11.0Equations of motion by graphical method

Figure below represents a velocity-time graph, in which $AB$ represents the initial velocity $u$, CE represents final velocity $v$, such that the change in velocity is represented by CD, which takes place in time $t$, represented by AE (see figure).

An object moving with certain initial velocity $u$ and attains final velocity $v$ after time $t$. Derivation of $\mathbf{v}=\mathbf{u}+$ at Acceleration = slope of the graph line BC $a=\frac{CD}{BD}=\frac{CE-DE}{BD}$ or $a=\frac{v-u}{t}$ $\left[\begin{array}{l}\because \mathrm{DE}=\mathrm{AB}=\mathrm{u}\\ \because \mathrm{BD}=\mathrm{AE}=\mathrm{t}\\ \because \mathrm{CE}=\mathrm{v}\end{array}\right]$ $\mathrm{v}-\mathrm{u}=\mathrm{at}$ $\mathbf{v}=\mathbf{u}+\mathbf{at}$

Derivation of s=u t+\frac{1}{2} at ^{2}

Distance travelled $=$ Area of rectangle ABDE + Area of triangle $BCD$ $=AB\times AE+\frac{1}{2}(BD\times CD)=u\times t+\frac{1}{2}[t\times (v-u)]$ $\left[\begin{array}{l}\because \mathrm{BD}=\mathrm{AE}=\mathrm{t}\\ \because \mathrm{CD}=\mathrm{CE}-\mathrm{DE}=\mathrm{v}-\mathrm{u}\end{array}\right]$ $=u\times t+\frac{1}{2}[t\times (u+at-u)][\because v=u+at]$ $s=ut+\frac{1}{2}{a}^2$

Derivation of \mathbf{v}^{2}=\mathbf{u}^{2}+2 as

From the velocity-time graph, distance covered = Area of trapezium ABCE $\Rightarrow s=\frac{1}{2}(AB+CE)\times AE\therefore s=\frac{1}{2}(u+v)\times t$

Acceleration $=\frac{\text{ Change in velocity }}{\text{ Time }}$ $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}$ $\therefore \mathrm{t}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}$

Substituting the value of $t$ in equation (i) $s=\frac{(v+u)}{2}\times \frac{(v-u)}{a}\left[\because A^2-B^2=(A+B)\times (A-B)\right]$ $\mathrm{s}=\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{a}}$, or $v^2-u^2=2$ as, or ${\mathbf{v}}^2={\mathbf{u}}^2+2as$

• Area of rectangle $=$ length $\times$ breadth
• Area of triangle $=\frac{1}{2}\times$ base $\times$ height
• Area of trapezium $=\frac{1}{2}\times$ sum of length of parallel sides $\times$ height or distance between parallel sides. $\left[\frac{1}{2}(a+b)\times h\right]$

Distance-time graph	Velocity-time graph
s-t graph for a body at rest
s-t graph for a body in uniform motion $(a=0)$	v-t graph for an object moving with uniform acceleration

• The slope of distance/displacement time graph gives the speed/velocity of the body.
• The slope of speed/velocity time graph gives the acceleration of the body.
• More the slope of distance/displacement time graph more will be the speed/velocity of the moving body.

Numerical Ability

• The distance covered by moving body was recorded after every $1\mathbf{s}$ for 4 s . The data is given in the table below. Represent it graphically and find the speed of the body from the graph.

Solution: Decode the problem Identify the graph According to data, body covers equal distance in equal interval of time. So graph will be distance time graph of uniform motion. Apply the formula $\text{Speed = Slope }=\frac{\text{ measure on }y\text{-axis }}{\text{ measure on }x\text{-axis }}$ The body covers equal distance in equal interval of time. The distance-time graph of a body in uniform motion is a straight line. The horizontal axis represents time, while the vertical axis represents distance (see figure).

If we study the graph carefully, we will see that $\mathrm{BE}/\mathrm{AE}$ would also give the speed of the body. BE gives the distance travelled in the time interval AE , which is 1 s .

• A car travels with a uniform velocity of $20{\mathbf{ms}}^{-1}$ for 5 s . The brakes are then applied and the car is uniformly retarded. It comes to rest in further 8 s . Draw a graph of velocity against time. Use this graph to find : (i) the distance travelled in first 5 s , (ii) the distance travelled after the brakes are applied, Solution:
  
  Decode the problem Draw the graph Take velocity on $y$-axis and time of x -axis. When the car travels with a uniform velocity, the graph is a straight line graph parallel to x -axis. When the car is uniformly retarded, the velocity decreases and becomes zero. The graph for this is a straight line with a decreasing slope. Identify the formula The area under the curve in a velocity time graph gives the distance travelled by the car. (i) The distance travelled in first 5 seconds $=$ area of rectangle OABC area of rectangle $=$ length $\times$ breadth $=20\mathrm{m}/\mathrm{s}\times 5=100\mathbf{m}$ (ii) The distance travelled in last 8 seconds (from 5 sec to 13 sec ) = area of the triangle $BCD=\frac{1}{2}\times$ base $\times$ height $=\frac{1}{2}\times (13-5)\mathrm{s}\times 20\mathrm{m}/\mathrm{s}$ $=\frac{1}{2}\times 8\mathrm{s}\times 20\mathrm{m}/\mathrm{s}=80\mathbf{m}$
• In the previous question, find the acceleration during the first $5\mathbf{s}$ and last $8s$. Solution: Decode the problem Study the graph (i) For first 5 seconds initial and final velocity is same. (ii) For the last 8 seconds, $u=20\mathrm{m}/\mathrm{s}$ and $v=0$. Identify the formula $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}$ Acceleration during first 5 second is zero as velocity is constant. $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{20-20}{5}$ $=0\mathrm{m}/{\mathrm{s}}^2$ For the last 8 seconds, the initial velocity of the car is $20\mathrm{m}/\mathrm{s}$ and the final velocity is 0 . In the last 8 seconds, acceleration, $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{0-20}{13-5}=\frac{20}{8}$ $=-2.5\mathrm{m}/{\mathrm{s}}^2$
• Study the speed-time graph of a body given here and answer the following questions.
  
  (a) What type of motion is represented by OA? (b) What type of motion is represented by AB? (c) What type of motion is represented by BC? Solution: Decode the problem Study the graph (a) Graph represents object moving with uniform acceleration (b) Graph represents object is in uniform motion (c) Graph represent object moving with uniform retardation negative acceleration. The $y$-axis of the graph denotes speed and the $x$-axis denotes time. (a) $OA$ is a straight line graph between speed and time. The slope of speed-time graph represent acceleration. Thus OA represent uniform acceleration. (b) AB is a straight line parallel to the time axis. The speed is constant from A to B . Thus AB represent uniform speed. (c) $BC$ is a straight-line graph between speed and time (downward slope). BC represents uniform retardation or negative acceleration.
• In the previous question, find the acceleration, retardation and the distance travelled by the body. Solutions: Decode the problem Study the graph (i) In first 4 seconds speed increases 0 to $6\mathrm{m}/\mathrm{s}$. (ii) In the last 6 seconds speed decreases from $6\mathrm{m}/\mathrm{s}$ to 0 . (iii) The area under the curve of a speed time graph gives the distance travelled by the body. Identify the formula (i) Acceleration = slope of line OA $\mathrm{a}=\frac{{\mathrm{v}}_2-{\mathrm{v}}_1}{{\mathrm{t}}_2-{\mathrm{t}}_1}$ (ii) Acceleration = slope of line BC $a=\frac{v_2-v_1}{t_2-t_1}$ (i) $\mathrm{a}=\frac{{\mathrm{v}}_2-{\mathrm{v}}_1}{{\mathrm{t}}_2-{\mathrm{t}}_1}=\frac{6-0}{4-0}$ $=1.5\mathrm{m}/{\mathrm{s}}^2$ (ii) $a=\frac{v_2-v_1}{t_2-t_1}=\frac{0-6}{16-10}$ $=-1\mathrm{m}/{\mathrm{s}}^2$ (iii) Distance travelled by the body = area of shaded portion $=$ area of triangle OAD + area of rectangle DABE + area of triangle BEC

12.0Circular motion

When a particle moves along a circular path, its motion is called 'circular motion' (see figure).

Circular motion of a particle

• In a circular motion, velocity of a particle is tangential to the circular path. If the body covers equal distances along the circumference of the circle, in equal intervals of time, then motion is said to be a uniform circular motion. When a body moves along a circular path, then its direction of motion changes continuously. Thus, a circular motion is always a non-uniform motion.
• A uniform circular motion is a motion in which speed remains constant but direction of velocity changes continuously.

Examples of uniform circular motion

(i) An athlete running on a circular track with constant speed. (ii) Motion of tips of the second hand, minute hand and hour hand of a wrist watch.

13.0Concept Map

Important terms, formula, unit, quantity