Kinematics Equations

Kinematics is a branch of physics that focuses on describing how objects move without considering the forces that cause this motion. It deals with key quantities such as displacement, velocity, acceleration, and time. When an object moves with constant acceleration, its motion can be predicted using the kinematic equations. These equations establish relationships between initial velocity, final velocity, acceleration, time, and displacement, allowing us to solve a wide range of motion problems; equations are essential tools in physics for analyzing straight-line motion. They help describe situations such as objects falling under gravity or vehicles speeding up and slowing down. By applying these equations correctly, we gain a clearer understanding of how motion changes over time and can interpret real-world motion more accurately.

Kinematics Equations for Uniformly Accelerated Motion

Consider an object undergoing uniformly accelerated motion with an acceleration 'a' along a straight line OX, starting from the origin O.

1.0Velocity-Time Relationship

Acceleration of the object = Change in velocity / Time taken

$[a = \frac{v _{2} - v _{1}}{t _{2} - t _{1}}]$

$[v_{2} - v_{1} = a (t_{2} - t_{1})]$

If

$[v_{1} = u, t_{1} = 0, v_{2} = v, t_{2} = t]$

Then

$[v = u + a t]$

2.0Position-Time Relationship

Average velocity

$[u_{a v} = \frac{x _{2} - x _{1}}{t _{2} - t _{1}}]$

$[u_{a v} = \frac{v _{1} + v _{2}}{2}]$

$[x_{2} - x_{1} = \frac{v _{1} + v _{2}}{2} (t_{2} - t_{1})]$

Substituting $(v_{2} = v_{1} + a (t_{2} - t_{1}))$ :

$[x_{2} = x_{1} + v_{1} (t_{2} - t_{1}) + \frac{1}{2} a (t_{2} - t_{1})^{2}]$

If

$[x_{1} = x_{0}, t_{1} = 0, v_{1} = u]$

Then

$[x - x_{0} = u t + \frac{1}{2} a t^{2}]$

$[s = u t + \frac{1}{2} a t^{2}]$

3.0Position-Velocity Relationship

From

$[v_{2} - v_{1} = a (t_{2} - t_{1})]$

$[t_{2} - t_{1} = \frac{v _{2} - v _{1}}{a}]$

Substitute into

$[x_{2} - x_{1} = \frac{v _{1} + v _{2}}{2} (t_{2} - t_{1})]$

$[x_{2} - x_{1} = \frac{v _{1} + v _{2}}{2} \cdot \frac{v _{2} - v _{1}}{a}]$

$[x_{2} - x_{1} = \frac{v _{2}^{2} - v _{1}^{2}}{2 a}]$

$[v_{2}^{2} - v_{1}^{2} = 2 a (x_{2} - x_{1})]$

If

$[v_{1} = u, x_{2} - x_{1} = s]$

Then

$[v^{2} - u^{2} = 2 a s]$

4.0Distance Travelled in the nth Second

$[D_{n} = S_{n} - S_{n - 1}]$

$[S = u t + \frac{1}{2} a t^{2}]$

$[S_{n - 1} = u (n - 1) + \frac{1}{2} a (n - 1)^{2}]$

$[D_{n} = [u n + \frac{1}{2} a n^{2}] - [u (n - 1) + \frac{1}{2} a (n - 1)^{2}]]$

$[D_{n} = u + \frac{a}{2} (2 n - 1)]$

Kinematics Equations for Uniformly Accelerated Motion (Graphical Method)

Acceleration = slope of velocity-time graph

$[a = \frac{v - u}{t}]$

$[v = u + a t]$

Distance travelled in time ( t ):

$[S = area under velocity-time graph]$

$[S = u t + \frac{1}{2} a t^{2}]$

Using trapezium method:

$[S = \frac{1}{2} (u + v) t]$

$[v^{2} - u^{2} = 2 a s]$

Kinematics Equations for Uniformly Accelerated Motion (Calculus Method)

5.0Velocity-Time Relationship

$[a = \frac{d v}{d t}]$

$[d v = a, d t]$

$[\int_{u}^{v} d v = \int_{0}^{t} a, d t]$

$[v - u = a t]$

$[v = u + a t]$

6.0Distance-Time Relationship

$[v = \frac{d x}{d t}]$

$[d x = v, d t]$

$[d x = (u + a t), d t]$

$[\int_{x_{0}}^{x} d x = \int_{0}^{t} (u + a t), d t]$

$[x - x_{0} = u t + \frac{1}{2} a t^{2}]$

$[s = u t + \frac{1}{2} a t^{2}]$

7.0Velocity-Displacement Relationship

$[a = \frac{d v}{d t} = \frac{d v}{d x} \cdot \frac{d x}{d t}]$

$[a = v \frac{d v}{d x}]$

$[a, d x = v, d v]$

$[\int_{x_{0}}^{x} a, d x = \int_{u}^{v} v, d v]$

$[a (x - x_{0}) = \frac{v ^{2} - u ^{2}}{2}]$

$[v^{2} - u^{2} = 2 a s]$

Illustration-1 Two particles A and B are situated at the same place initially. Particle A moves with constant speed 20 m/s while B starts moving with constant acceleration 5m/s2 in the same direction. Find where and when they will cross each other.

Solution:

$[s_{1} = s_{2}]$

$[u_{1} t + \frac{1}{2} a_{1} t^{2} = u_{2} t + \frac{1}{2} a_{2} t^{2}]$

$[20 t + \frac{1}{2} (0) t^{2} = 0 t + \frac{1}{2} (5) t^{2}]$

$[20 t = \frac{5}{2} t^{2}]$

$[t^{2} - 8 t = 0]$

$[t = 0, t = 8]$

$[s_{A} = 20 \times 8 = 160 m]$

Illustration-2 A police inspector in a jeep is chasing a pickpocket on a straight road. The jeep is going at its maximum speed v (assumed uniform). The pickpocket rides on the motorcycle of a waiting friend when the jeep is at

a distance d away, and the motorcycle starts with a constant acceleration a. Show that the pick pocket will be caught if v2ad .

Solution: Suppose the pickpocket is caught at a time t after the motorcycle starts. The distance travelled by the motorcycle during this interval is

Distance travelled by motorcycle:

$[s = \frac{1}{2} a t^{2}]$

Distance travelled by jeep:

$[s + d = v t]$

$[\frac{1}{2} a t^{2} + d = v t]$

$[t = \frac{v \pm v ^{2} - 2 a d}{a}]$

For real positive $(t)$ :

$[v^{2} \geq 2 a d]$

Kinematics Equations

Kinematics Equations for Uniformly Accelerated Motion

1.0Velocity-Time Relationship

2.0Position-Time Relationship

3.0Position-Velocity Relationship

4.0Distance Travelled in the nth Second

Kinematics Equations for Uniformly Accelerated Motion (Graphical Method)

Kinematics Equations for Uniformly Accelerated Motion (Calculus Method)

5.0Velocity-Time Relationship

6.0Distance-Time Relationship

7.0Velocity-Displacement Relationship

Table of Contents

Frequently Asked Questions

Why can kinematic equations be used only for constant acceleration?

How is displacement different from distance in kinematics?

What does a negative value of acceleration indicate in kinematic equations?

Can an object have zero displacement but non-zero distance?

Why does the equation for distance in the nth second help in uniformly accelerated motion?

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