Kinematics Previous Year Questions With Solutions

1.0Introduction

Kinematics is the study of motion—how objects move in terms of position, velocity, and acceleration—without looking at the causes of the motion (like forces). It includes both linear motion (straight-line paths) and rotational motion (around a point or axis). The main variables in kinematics are displacement (change in position), velocity (speed with direction), acceleration (change in velocity), and time. Using kinematic equations, we can predict future motion or understand past movement based on known values.

2.0Key Concepts to Remember

Displacement in Vector Form

Displacement vector $\overrightarrow{\Delta r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

Sense of Direction in terms of base vectors

Galileo’s law of odd numbers.

• When particle starts from rest and moves with constant acceleration then ratio of distance travelled by it in successive equal intervals of time is

1 : 3 : 5 : 7........ (2n – 1)

Stopping Distance

• The distance a vehicle travels after braking before it stops is called the stopping distance.

$s=\frac{u_0^2}{2a_0}[\text{since a is constant}]\Rightarrow s\propto u_0^2$

if u becomes n times then s becomes n2 times that of previous value.

Stopping Time

$t=\frac{u_0}{a_0}[\text{since a is constant}]\Rightarrow t\propto u_0$

if u becomes n times then t becomes n times that of previous value.

Reaction Time

• When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act.

 Motion Under Gravity

• The acceleration a body experiences due to gravity is called acceleration due to gravity, denoted by g.

$g=9.8{\text{m/s}}^2\text{or}g=980{\text{cm/s}}^2\text{or}g=32{\text{ft/s}}^2$

Sign Conventions

Note: Negative and positive signs are matters of our choice, so we can select any direction as positive and opposite side as negative.

3.0Projectile Motion

• When a body moves with constant acceleration such that its initial velocity and acceleration are non- collinear then its path is parabola and motion is known as projectile motion.

4.0River-Boat (or Man) Problem

5.0Important Formulas

Displacement vector	$\overrightarrow{\Delta r}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$
Average Speed	$V_{avg}=\frac{v_1+v_2+\dots +v_n}{n}$
Time Average Velocity	$\text{If }v=f(t)\Rightarrow <v>=\frac{{\int }_{t_1}^{t_2}vdt}{{\int }_{t_1}^{t_2}dt}$
Space Average Velocity	$\text{If }v=f(x)\Rightarrow <v>=\frac{{\int }_{x_1}^{x_2}vdx}{{\int }_{x_1}^{x_2}dx}$
Instantaneous Velocity	${\vec{v}}_{\text{inst}}={\lim }_{\Delta t\to 0}\left(\frac{\Delta \vec{r}}{\Delta t}\right)\Rightarrow {\vec{v}}_{\text{inst}}=\frac{d\vec{r}}{dt}$
Change in velocity	$|\Delta \vec{v}|=2v\sin \left(\frac{\theta }{2}\right)$
Instantaneous Acceleration	${\vec{a}}_{\text{inst}}={\lim }_{\Delta t\to 0}\left(\frac{\Delta \vec{v}}{\Delta t}\right)\Rightarrow {\vec{a}}_{\text{inst}}=\left(\frac{d\vec{v}}{dt}\right)$
Equations of Motion	$v=u+at$ $s=ut+\frac{1}{2}a{t}^2$ $v^2=u^2+2as$ $S_{n^{th}}=u+\frac{a}{2}(2n-1)$
Stopping Distance	$s=\frac{u_0^2}{2a_0}[\text{since a is constant}]\Rightarrow s\propto u_0^2$
Stopping Time	$t=\frac{u_0}{a_0}[\text{since a is constant}]\Rightarrow t\propto u_0$
Horizontal Motion	$v_{\text{max}}=\left(\frac{\alpha \beta }{\alpha +\beta }\right)T$ $S=\frac{1}{2}\left(\frac{\alpha \beta }{\alpha +\beta }\right)T^2$
Maximum Height (H)	$H=\frac{u^2}{2g}$
Total Time of Flight (T)	$T=\frac{2u}{g}$
Equation of Trajectory	$y=x\tan \theta \left[1-\frac{x}{R}\right]$
Kinetic energy at highest point	$K=\frac{1}{2}m(u\cos \theta {)}^2=K_o{\cos }^2\theta$
Relative Displacement	$x_{BA}=x_B-x_A=\frac{d{x}_B}{dt}-\frac{d{x}_A}{dt}={\vec{v}}_B-{\vec{v}}_A$ $\Rightarrow {\vec{v}}_{BA}={\vec{v}}_B-{\vec{v}}_A$
Equations of Motion (Relative)	$v_{\text{rel}}=u_{\text{rel}}+a_{\text{rel}}t$ $s_{\text{rel}}=u_{\text{rel}}t+\frac{1}{2}{a}_{\text{rel}}{t}^2$ $v_{\text{rel}}^2=u_{\text{rel}}^2+2a_{\text{rel}}{s}_{\text{rel}}$ $s_{\text{rel}}=\frac{1}{2}(u_{\text{rel}}+v_{\text{rel}})t$
Relative velocity in a plane	${\vec{v}}_{AB}={\vec{v}}_A-{\vec{v}}_B$ $|{\vec{v}}_{AB}|=\sqrt{v_A^2+v_B^2-2v_A{v}_B\cos \theta }$
Rain-Man Concept	$v_{RM}=\sqrt{v_R^2+v_M^2}$ $\theta ={\tan }^{-1}\left(\frac{v_M}{v_R}\right)$
River-Boat (or Man) Problem	$T=\frac{d}{v\cos \theta }=\frac{d}{\sqrt{v^2-v_M^2}},\text{drift of man:}{t}_{\text{min}}=\frac{d}{v_R}$

6.0Past Year Questions with Solutions on Kinematics: JEE (Mains)

Q-1.Position of an ant (S in metres) moving in Y-Z plane is given by  $S=2t^2\hat{j}+5\hat{k}$ (where t is in second). The magnitude and direction of velocity of the ant at t = 1 s will be :

$(1)16\frac{m}{s}iny-direction$

$(2)4\frac{m}{s}inx-direction$

$(3)9\frac{m}{s}inz-direction$

$(4)4\frac{m}{s}iny-direction$

Solution: Ans(4)

$\vec{v}=\frac{d\vec{s}}{dt}=4t\hat{j}$

$\text{at }t=1\Rightarrow \vec{v}=4\hat{j}$

Q-2. A particle starts from origin at t = 0 with a velocity 5 i m/s  and moves in x-y  plane under action of a force which produces a constant acceleration of $3i+2jm/s2(3\hat{i}+2\hat{j}){\text{m/s}}^2$ . If the x- coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is $\sqrt{a}\text{m/s}$. The value of is ______.

Solution: Ans $\sqrt{673}\text{m/s}$

$u_x=5\text{m/s},a_x=3{\text{m/s}}^2,x=84\text{m}$

$v_x^2-u_x^2=2ax$

$v_x^2-25=2(3)(84)\Rightarrow v_x=23\text{m/s}$

$v_y-u_y=a_{y}t$

$t=\frac{23-5}{3}=6\text{s}$

$v_y=0+a_{y}t=0+2\times (6)=12\text{m/s}$

$v^2=v_x^2+v_y^2=23^2+12^2=673\Rightarrow v=\sqrt{673}\text{m/s}$

Q-3. A bullet fired into a fixed target loses one third of its velocity after travelling4 cm. It penetrates  further D × 10^–3 m before coming to rest.  The value of D is :

(1)2    (2)5    (3) 3    (4) 4

Solution:Ans(Bonus)

$v^2-u^2=2as$

${\left(\frac{2u}{3}\right)}^2=u^2+2(-a)(4\times 10^{-2})$

$\frac{4u^2}{9}=u^2-2a(4\times 10^{-2})$

$-\frac{5u^2}{9}=-2a(4\times 10^{-2})\cdots \text{(1)}$

$0={\left(\frac{2u}{3}\right)}^2+2(-a)x$

$\Rightarrow \frac{-4u^2}{9}=-2ax\cdots \text{(2)}$

12 we get

$\frac{5}{4}=\frac{4\times 10^{-2}}{x}$

$x=\frac{16}{5}\times 10^{-2}=3.2\times 10^{-2}=32\times 10^{-3}\text{m}$

Note: Since no option is matching

Q-4.A body falling under gravity covers two points A and B separated by 80 m in 2s. The distance of upper point A from the starting point is ________ m (use g = 10 ms^–2)

Solution: Ans(45) 

From AB

$$\begin{gathered} -80=-v_{1}t-\frac{1}{2}\times 10\times t^2 \\ -80=-2v_1-\frac{1}{2}\times 10\times 2^2 \\ -80=-2v_1-20 \\ -60=-2v_1\Rightarrow v_1=30\text{m/s} \\ \text{From O to A} \\ {v}^2=u^2+2gs \\ 30^2=0+2\times (-10)(-s) \\ 900=20s\to s=45m \end{gathered}$$

Q-5.A body starts moving from rest with constant acceleration covering displacement S1 in first (p – 1)seconds and S2 in the first p  seconds. The displacement S1+ S2 will be made in time :

$(1)(2p+1)s(2)\sqrt{(2p^2-p+1)s}(3)(2p-1)s(4)(2p^2-p+1)s$

Solution: Ans(2)

S1 in first (p-1) sec

S2 in first p sec

$$\begin{gathered} S_1=\frac{1}{2}a(p-1{)}^2 \\ {S}_2=\frac{1}{2}a(p{)}^2 \\ {S}_1+S_2=\frac{1}{2}a{t}^2 \\ (p-1{)}^2+p^2=t^2 \\ t=\sqrt{(2p^2-p+1)s} \end{gathered}$$

Q-6.A ball rolls off the top of a stairway with horizontal velocity u.The steps are 0.1 m high and 0.1 m wide. The minimum velocity u with which that ball just hits the step 5 of the stairway will be $\sqrt{x}\text{m/s}$ where x = ___________

[use g = 10 m/s2].

Solution: Ans(2)

The ball needs to just cross 4 steps to just hit 5th step

Therefore,horizontal Range R=0.4m

R=ut

Similarly in vertical direction

$$\begin{gathered} h=\frac{1}{2}g{t}^2\Rightarrow 0.4=\frac{1}{2}g{t}^2 \\ 0.4=\frac{1}{2}g{\left(\frac{0.4}{u}\right)}^2 \\ {u}^2=2\Rightarrow u=\sqrt{2}\text{m/s} \\ Thereforex=2 \end{gathered}$$

Q-7.A particle is moving in a straight line. The variation of position ‘x’  as a function of time ‘t’ is given as $x=(t^3-6t^2+20t+15)\text{m}$.  The velocity of the body when its acceleration becomes zero is :

$(1)4\frac{m}{s}(2)8\frac{m}{s}(3)10\frac{m}{s}(4)6\frac{m}{s}$

Solution: Ans(2)

$$\begin{gathered} x=(t^3-6t^2+20t+15) \\ \frac{dx}{dt}=v=3t^2-12t+20 \\ \frac{dv}{dt}=a=6t-12 \\ Whena=0 \\ 6t-12=0t=2sec \\ \text{At t= 2 sec} \\ v=3(2{)}^2-12(2)+20\Rightarrow v=8\text{m/s} \end{gathered}$$

Q-8.A particle of mass m  projected with a velocity ‘u’making an angle of 30º with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is : 

$(1)\frac{\sqrt{3}}{16}\cdot \frac{m{u}^3}{g}(2)\frac{\sqrt{3}}{2}\cdot \frac{m{u}^2}{g}(3)\frac{m{u}^3}{\sqrt{2}g}(4)\text{Zero}$

Solution: Ans(1)

$L=mu\cos \theta \cdot H$

$$\begin{gathered} L=mu\cos \theta \times \frac{u^2{\sin }^2\theta }{2}g \\ L=\frac{m{u}^3}{2g}\times \frac{\sqrt{3}}{2}\times {\left(\frac{1}{2}\right)}^2 \\ =\frac{\sqrt{3}}{16}\cdot \frac{m{u}^3}{g} \end{gathered}$$

Q-9.The displacement and the increase in the velocity of a moving particle in the time interval of t to (t + 1) s are 125 m and 50 m/s, respectively. The distance travelled by the particle in (t + 2)th s  is ____ m.

Solution: Ans(175)

Considering acceleration is constant

$$\begin{gathered} v=u+at \\ u+50=u+a\to a=50/fracm{s}^2 \\ 125=ut+\frac{1}{2}a{t}^2 \\ 125=u+\frac{a}{2} \\ \to u=100\frac{m}{s} \\ {S}_{n^{\text{th}}}=u+\frac{a}{2}[2n-1]=100+\frac{50}{2}[2\times 2-1] \\ 100+253=175m \end{gathered}$$

Q-10.Projectiles A and B are thrown at angles of 45° and 60°with vertical respectively from top of a 400 m high tower. If their ranges and times of flight are same, the ratio of their speeds of projection vA : vB is :

$(1)1:\sqrt{3}(2)\sqrt{2}:1(3)1:2(4)1:\sqrt{2}$

Solution: Ans(Bonus)

For $u_{A}and{u}_B$ time of flight and range cannot be same

Q-11. The relation between time ‘t’ and distance ‘x’is $t=a{x}^2+bx,$ where a and b are constants. The relation between acceleration (a) and velocity (v) is:

$1)a=-2a{v}^{3}2)a=-5a{v}^{5}3)a=-3a{v}^{2}4)a=-4a{v}^4$

Solution: Ans(1)

$$\begin{gathered} t=\alpha x^2+\beta x \\ \frac{dt}{dx}=2\alpha x+\beta \\ \frac{1}{v}=2\alpha x+\beta \\ -\frac{1}{v^2}\cdot \frac{dv}{dt}=2\alpha \cdot \frac{dx}{dt} \\ \frac{dv}{dt}=-2\alpha v^3 \end{gathered}$$

Q-12.A body starts falling freely from height H hits an inclined plane in its path at height h. As a result of this perfectly elastic impact, the direction of the velocity of the body becomes horizontal. The value of $\frac{H}{h}$ for which the body will take the maximum time to reach the ground is _____. 

Solution:Ans(2)

Total time of flight=T

$$\begin{gathered} T=\sqrt{\frac{2h}{g}}+\sqrt{\frac{2(H-h)}{g}} \\ \frac{dT}{dh}=0 \\ \sqrt{\frac{2}{g}}\left(\frac{-1}{2\sqrt{H-h}}+\frac{1}{2\sqrt{h}}\right)=0 \\ \sqrt{H-h}=\sqrt{h} \\ \Rightarrow h=\frac{H}{2}\Rightarrow \frac{H}{h}=2 \end{gathered}$$

Q-13.A particle moving in a circle of radius R with uniform speed takes time T  to complete one revolution. If this particle is projected with the same speed at an angle to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection   is then given by :

$(1){\sin }^{-1}\left[{\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)}^{\frac{1}{2}}\right](2){\sin }^{-1}\left[{\left(\frac{\pi R}{2g{T}^2}\right)}^{\frac{1}{2}}\right](3){\cos }^{-1}\left[{\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)}^{\frac{1}{2}}\right](4){\cos }^{-1}\left[{\left(\frac{\pi R}{2g{T}^2}\right)}^{\frac{1}{2}}\right]$

Solution:Ans(1)

$$\begin{gathered} \frac{2\pi R}{T}=V \\ \text{Maximum Height, }H=\frac{v^2{\sin }^2\theta }{2g} \\ 4R=\frac{4{\pi }^2{R}^2}{T^{2}2g}{\sin }^2\theta \\ \sin \theta =\sqrt{\frac{2g{T}^2}{{\pi }^{2}R}} \\ \theta ={\sin }^{-1}\left({\left(\frac{2g{T}^2}{{\pi }^{2}R}\right)}^{\frac{1}{2}}\right) \end{gathered}$$

Q-14.A particle is moving in one dimension (along x axis) under the action of a variable force. Its initial position was 16 m right of origin. The variation of its position (x)  with time (t) is given as $x=-3t^3+18t^2+16t$ , where x is in m and t is in s. The velocity of the particle when its acceleration becomes zero is _________ m/s.

Solution: Ans(52)

$$\begin{gathered} x=-3t^3+18t^2+16t \\ v=\frac{dx}{dt}=-9t^2+36t+16 \\ a=-18t+36 \\ a=0att=2s \\ v=-9(2{)}^2+36\times 2+16\Rightarrow v=52\text{m/s} \end{gathered}$$

Q-15.Train A  is moving along two parallel rail tracks towards north with speed 72 km/h and train B is moving towards south with speed108 km/h . Velocity of train B  with respect to A and velocity of ground with respect to B are (in ms^–1) :

(1 ) –30 and 50          (2) –50 and –30          (3)   –50 and 30          (4)  50 and –30  

Solution: Ans(3)

$v_A=20\text{m/s},v_B=-30\text{m/s}$

$$\begin{gathered} \text{Velocity of B w.r.t A:}{v}_{B/A}=-50\text{m/s} \\ \text{Velocity of Ground w.r.t B:}{v}_{G/B}=30\text{m/s} \end{gathered}$$

Q-16.A particle initially at rest starts moving from reference point. x = 0 alongx-axis , with velocity v that varies as $v=4\sqrt{x}\frac{m}{s}$ .The acceleration of the particle is ____ $m{s}^{-2}$.

Solution:Ans(8)

$$\begin{gathered} v=4\sqrt{x} \\ a=v\frac{dv}{dx}=4\sqrt{x}\times 4\times \frac{1}{2}{x}^{-1/2}=8{\text{m/s}}^2 \end{gathered}$$