Kinematics holds a weightage of 3-4% with 1-2 questions that are asked in JEE Main exam. Questions from Kinematics are usually direct and therefore scoring. To score well in this section, students need to be thorough with knowledge of formulas, dimensions and graphs. Check JEE Main Sample Papers

Important topics from Kinematics in JEE Main include Frame of reference, Motion in a straight line, Position-time graph, speed and velocity, Uniform and non-uniform motion, average speed and instantaneous velocity Uniformly accelerated motion, velocity-time, position-time graphs, relations for uniformly accelerated motion, Scalars and Vectors, Vector addition and Subtraction, Zero Vector, Scalar and Vector products, Unit Vector, Resolution of a Vector, Relative Velocity, Motion in a plane, Projectile Motion, Uniform Circular Motion. Check JEE Main Physics Syllabus

Kinematics is a part of physics which deals with motion and position of objects, points, and systems. Through kinematics, one can find lots of answers related to objects in motion, robotics, etc. The motion of an object could either be linear or circular as well as one dimensional, two dimensional, or three dimensional.

Questions on JEE Main Kinematics are usually based on speed, velocity, distance, acceleration, and motion in a straight line. The article below covers previous year questions and solutions on Kinematics and a few quick formulas for revision. Check JEE Main Physics Preparation

JEE Main Kinematics Previous Year Questions and Solutions

1. A small block starts sliding down without friction from an inclined plane. Consider Sn to be the distance travelled from time t = (n – 1) to t = (n). Then Sn/ Sn + 1 is? [2 Marks – 2004 (Screening)]
   1. 2n - 12n
   2. 2n + 12n - 1
   3. 2n - 12n + 1
   4. 2n2n + 1

Answer: Option (c) 2n - 12n + 1

Solution:

Distance Travelled in tth second is St

• Given that u = 0
• St = u + at - a2
• SnSn + 1 = an - a2 a n+1 - a2 = a 2n - 1 × 22a 2n +1 = 2n - 1 2n + 1

2. From a tower of Height H, a particle is thrown upwards vertically, with a speed u. Time taken by the particle to hit the ground is n times that taken for it to reach the highest point of its path.

What is the relation between H, u, and n? [4 Marks – 2014]

• 1. g H = u2 (n – 2)
  2. 2g H = nu2 (n – 2)
  3. 2g H = n2 u2
  4. g H = u2 (n – 2)2

Answer: Option (b) 2g H = nu2 (n – 2)

Solution:

Time to reach the highest point of its path is t = ug

Time to reach ground = nt

• S = ut + 12 at2
• -H = u (nt) - 12 g (nt)2
• 2g H = nu2 (n – 2)

3. A ball with Kinetic Energy E, is projected at an angle of 45˚ to the horizontal. The kinetic energy of the ball at highest point of its height will be: [2002]
   1. E/√2
   2. E
   3. E/2
   4. Zero

Answer: Option (c) E/2

Solution:

Let mass of ball be m and projected speed be u.

So, the Kinetic Energy E at the point of projection = ½ mu2

At the highest point vertical component velocity = 0 and

Horizontal component velocity = u cosϴ = u cos 45˚

At the highest point velocity = u cosϴ = u cos 45˚ = u/√2

• Kinetic energy at highest point = ½ m (u/√2)2
• = ½ mu2 × ½ = E/2

4. A particle of mass m moves on x-axis as follows: It starts from rest at t = 0, from point x = 0 and comes to rest at t = 1 at the point x = 1. No other information is available about its motion at intermediate times (0 < t < 1). If α denotes the instantaneous acceleration of the particle, then: [2 Marks – 1993]

(Question Type: More than one correct answer)

• 1. α cannot remain positive for all t in the interval 0 ≤ t ≤ 1
  2. α cannot exceed 2 at any point in its path
  3. α should be ≥ 4 at some point or points in its path
  4. α should change sign during the motion but no other assertion can be done with the information provide

Answer: Options a, c, d

Solution:

The body is at rest at t = 0 and t = 1.

Initially α is positive so that it can acquire velocity. Then α has to be negative so that body can come to rest. Therefore, α cannot remain positive for all time in the interval 0 ≤ t ≤ 1.

Hence, options (a) and (d) are correct.

Next, as shown in the graph, the journey is v – t.

Total time of journey is 1 sec

Total displacement = 1 m

i.e Area under (v – t) graph.

Vmax = 2st = 2 × 11 = 2 m/s

For path OB,

acceleration (α) = Change in Velocitytime = 21/2 = 4 m/s2

For path BD, retardation = -4 m/s2

For path OA, α (acceleration) > 4 m/s2

For path AD, α (retardation) < -4 m/s2

For path OC, acceleration α < 4 m/s2

For path CD, retardation α > 4 m/s2

Hence, at some point or points in its path α is ≥ 4. So option c is correct.

5. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, total area around the fountain that gets wet is: [4 Marks – 2011]
   1. π  V4g2
   2. π2  V4g2
   3. π  V2g2
   4. π  V2g

Answer: Option (a) π V4g2

Solution:

Maximum range of water coming out of the fountain,

Rmax =  V2sin2ϴg =  V2sin90˚g =  V2g

Total area of Foundation = π R2max = π  V4g2

6. A boy can throw a stone up to a maximum height of 10 metres. Maximum horizontal distance up to which the boy can throw the same stone will be [4 Marks – 2012]
   1. 20√2 m
   2. 10 m
   3. 10√2 m
   4. 20 m

Answer: Option (d) 20 m

Solution:

We know that,

R =  u2sin2ϴg

H =  u2sin2ϴ2g

Hmax is possible if ϴ = 90˚

Hmax =  u2g = 10

• u2 = 10g × 2
• As R =  u2sin2ϴg
• Range is maximum when projectile is thrown at angle of 45˚

Rmax =  u2g = 10 × g × 2g = 20 meter

7. Speed of two identical cars are u and 4u at a specific instant. Ratio of respective distances in which the two cars are stopped from that instant is [2002]
   1. 1:4
   2. 1:16
   3. 1:1
   4. 1:8

Answer: Option (b) 1:16

Solution:

Given that,

Initial speed of two cars is u and 4u

Final speed is v = 0 for both cars

Gradually both cars are slowing down.

Hence, acceleration = (-a) (-a)

• So formula becomes

0 = u2 – 2as

• For car 1: u2 = 2as1 … (1)
• For car 2: (4u)2 = 2as2 … (2)
• Dividing (1) and (2) we get,
•  u216u2 =  2as12as2
•  116 =  s1s2

8. A particle is projected at 60˚ to horizontal with a kinetic energy K. Kinetic energy at the highest point is: [4 Marks - 2007]
   1. K/2
   2. K
   3. K/4
   4. Zero

Answer: Option (c) K/4

Solution:

Consider u as the velocity with which the particle is thrown and m as the mass of the particle.

Then,

KE = ½ mu2 … (1)

At the highest point velocity = u cos 60˚ (Only horizontal component is left, vertical component will be zero at the top-most point)

Thus,

Kinetic energy at highest point = (KE)H = ½ mu2 cos260˚ = K/4 (from equation (1))

9. A car moving at a speed of 50 Km/hr can be stopped by brakes after at least 6 metres. If the same car moves at a speed of 100 Km/hr, the minimum stopping distance is: [2003]
   1. 12 m
   2. 24 m
   3. 18 m
   4. 6 m

Answer: Option (b) 24 m

Solution:

For case 1:

U = 50 Km/hr = 50 × 10003600 m/s = 1259 m/s

V=0, s = 6 m, a =?

02 = u2 – 2as

• a = -  u22s
• a = -  (1259)22 ×6 = - 16 m/s2

For case 2:

U = 100 Km/hr = 100 × 10003600 m/s = 2509 m/s

V=0, a = - 16 m/s2, s =?

02 = u2 – 2as

• s = -  u22a
• s = -  (2509)22 ×-16 = - 16 m/s2

10. Which of these statements is FALSE for a particle moving in a circle with constant angular speed? [4 Marks - 2007]
    1. The velocity vector is tangent to the circle
    2. The acceleration vector points to the centre of the circle
    3. The acceleration vector is tangent to the circle
    4. The velocity and acceleration vectors are perpendicular to each other.

Answer: Option (c) The acceleration vector is tangent to the circle

Solution:

Option c is false because acceleration vector acts along the radius of circle or towards the centre of circle for uniform circular motion whereas velocity vector always acts along the tangent of circle.

JEE Main Kinematics- Quick Formulas

Candidates need to remember the important formulas from JEE Main Physics syllabus, as some questions could be solved directly using those formulas. During preparations it is highly suggested to write down the formulas of each topic separately and revise them regularly. The list of the formulas from Kinematics chapter are as follows:

1. Speed = Distance/ Time
2. Velocity = Displacement/ Time
3. Average Speed = Total distance travelled / Total time taken
4. Average Velocity = Total displacement/ Time taken
5. Instantaneous Velocity = Vinst =  dsdt
6. Acceleration =  dvdt
7. Equations of Motion
   1. V = u + at
   2. S = ut + ½ at2
   3. 2as = v2 – u2
   4. Sn = u + a (n – ½)
8. Equations of Motion for freely falling body (u = 0 and a = g)
   1. V = gt
   2. S = ½ gt2
   3. 2gs = v2
   4. Sn = g (n – ½)
9. Equations of Motion for body thrown vertically upwards (s = h and a = - g)
   1. V = u - gt
   2. h = ut - ½ gt2
   3. v2 = u2 – 2gh
   4. hn = u - g (n – ½)

Also Check JEE Main Kinematics Study Notes