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
Answer: Option (c) 2n - 12n + 1
Solution:
Distance Travelled in tth second is St
What is the relation between H, u, and n? [4 Marks – 2014]
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
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
(Question Type: More than one correct answer)
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.
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
Answer: Option (d) 20 m
Solution:
We know that,
R = u2sin2ϴg
H = u2sin2ϴ2g
Hmax is possible if ϴ = 90˚
Hmax = u2g = 10
Rmax = u2g = 10 × g × 2g = 20 meter
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)
0 = u2 – 2as
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))
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
For case 2:
U = 100 Km/hr = 100 × 10003600 m/s = 2509 m/s
V=0, a = - 16 m/s2, s =?
02 = u2 – 2as
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.
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:
Also Check JEE Main Kinematics Study Notes
*The article might have information for the previous academic years, please refer the official website of the exam.