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.
JEE Main Previous Year Solved Questions on Kinematics
JEE Adv Previous Year Solved Questions on Kinematics
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=2a0u02[since a is constant]⇒s∝u02
if u becomes n times then s becomes n2 times that of previous value.
Stopping Time
t=a0u0[since a is constant]⇒t∝u0
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=9.8m/s2org=980cm/s2org=32ft/s2
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
Δr=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^
Average Speed
Vavg=nv1+v2+…+vn
Time Average Velocity
If v=f(t)⇒<v>=∫t1t2dt∫t1t2vdt
Space Average Velocity
If v=f(x)⇒<v>=∫x1x2dx∫x1x2vdx
Instantaneous Velocity
vinst=limΔt→0(ΔtΔr)⇒vinst=dtdr
Change in velocity
∣Δv∣=2vsin(2θ)
Instantaneous Acceleration
ainst=limΔt→0(ΔtΔv)⇒ainst=(dtdv)
Equations of Motion
v=u+at
s=ut+21at2
v2=u2+2as
Snth=u+2a(2n−1)
Stopping Distance
s=2a0u02[since a is constant]⇒s∝u02
Stopping Time
t=a0u0[since a is constant]⇒t∝u0
Horizontal Motion
vmax=(α+βαβ)T
S=21(α+βαβ)T2
Maximum Height (H)
H=2gu2
Total Time of Flight (T)
T=g2u
Equation of Trajectory
y=xtanθ[1−Rx]
Kinetic energy at highest point
K=21m(ucosθ)2=Kocos2θ
Relative Displacement
xBA=xB−xA=dtdxB−dtdxA=vB−vA
⇒vBA=vB−vA
Equations of Motion (Relative)
vrel=urel+arelt
srel=urelt+21arelt2
vrel2=urel2+2arelsrel
srel=21(urel+vrel)t
Relative velocity in a plane
vAB=vA−vB
∣vAB∣=vA2+vB2−2vAvBcosθ
Rain-Man Concept
vRM=vR2+vM2
θ=tan−1(vRvM)
River-Boat (or Man) Problem
T=vcosθd=v2−vM2d,drift of man:tmin=vRd
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=2t2j^+5k^ (where t is in second). The magnitude and direction of velocity of the ant at t = 1 s will be :
(1)16sminy−direction
(2)4sminx−direction
(3)9sminz−direction
(4)4sminy−direction
Solution: Ans(4)
v=dtds=4tj^
at t=1⇒v=4j^
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 (3i^+2j^)m/s2 . If the x- coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is am/s. The value of is ______.
Solution: Ans 673m/s
ux=5m/s,ax=3m/s2,x=84m
vx2−ux2=2ax
vx2−25=2(3)(84)⇒vx=23m/s
vy−uy=ayt
t=323−5=6s
vy=0+ayt=0+2×(6)=12m/s
v2=vx2+vy2=232+122=673⇒v=673m/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)
v2−u2=2as
(32u)2=u2+2(−a)(4×10−2)
94u2=u2−2a(4×10−2)
−95u2=−2a(4×10−2)⋯(1)
0=(32u)2+2(−a)x
⇒9−4u2=−2ax⋯(2)
12 we get
45=x4×10−2
x=516×10−2=3.2×10−2=32×10−3m
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
−80=−v1t−21×10×t2−80=−2v1−21×10×22−80=−2v1−20−60=−2v1⇒v1=30m/sFrom O to Av2=u2+2gs302=0+2×(−10)(−s)900=20s→s=45m
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 :
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 xm/s where x = ___________
[use g = 10 m/s2].
Solution: Ans(2)
The ball needs to just cross 4 steps to just hit 5th step
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=(t3−6t2+20t+15)m. The velocity of the body when its acceleration becomes zero is :
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 :
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.
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)1:3(2)2:1 (3)1:2 (4)1:2
Solution: Ans(Bonus)
For uAanduB time of flight and range cannot be same
Q-11. The relation between time ‘t’ and distance ‘x’is t=ax2+bx, where a and b are constants. The relation between acceleration (a) and velocity (v) is:
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 hH for which the body will take the maximum time to reach the ground is _____.
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 :
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=−3t3+18t2+16t , where x is in m and t is in s. The velocity of the particle when its acceleration becomes zero is _________ m/s.
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)
B↓30m/sA↑20m/s
vA=20m/s,vB=−30m/s
Velocity of B w.r.t A:vB/A=−50m/sVelocity of Ground w.r.t B:vG/B=30m/s
Q-16.A particle initially at rest starts moving from reference point. x = 0 alongx-axis , with velocity v that varies as v=4xsm .The acceleration of the particle is ____ ms−2.
Solution:Ans(8)
v=4xa=vdxdv=4x×4×21x−1/2=8m/s2
Table of Contents
1.0Introduction
2.0Key Concepts to Remember
3.0Projectile Motion
4.0River-Boat (or Man) Problem
5.0Important Formulas
6.0Past Year Questions with Solutions on Kinematics: JEE (Mains)