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Last updated date: 19th Mar 2023

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**Kinematics** is an important topic for JEE exams. It mainly consists of one-dimensional and two-dimensional motion, and the questions asked in the examination are based on their applications. Compared to other concepts, this section will become easier with adequate practice and conceptual understanding.

In kinematics, first, we will learn about motion along a straight line and discuss different physical quantities like distance, displacement, velocity, etc., and their formulas. **Different types of motions** are also discussed in this section. Problems related to the three kinematical equations of motion for uniformly accelerated motion are very important. We will also see how we can draw a conclusion about the motion by evaluating and finding required quantities from various graphs.

Coming to motion along a** plane, properties of vectors** and the law of vector addition like triangular law and parallelogram law of vector additions, scalar and vector product of vectors are important. The problems with projectile motion on a horizontal and inclined plane are frequently asked in JEE exam.

Let us see what is kinematics and the important concepts and formulas needed to score a full marks for this topic in JEE Main with the help of some solved kinematics examples.

Motion in a Straight Line

Kinematics Equations

Motion under gravity

Graphs for all types of motion

Motion in a Plane

Projectile Motion

Resolution of Vector Rectangular Components

Vector Addition

Circular Motion

1. A balloon carrying a stone rises from rest on the ground with an acceleration of 10 m/s2 in the upward direction. After 5 sec, the stone is released and ultimately reaches the ground. Find the maximum height attained by the height.

Ans: Here, we will take the upward direction as positive and the downward direction as negative. Given the balloon rises up from rest. Therefore,

The initial velocity of the balloon, u1 = 0 m/s

Acceleration of the ball, a = 10 m/s2

First of all, let us find the height reached by the balloon in 5 seconds before the stone is released using the equation of motion.

$s=ut+\dfrac{1}{2}at^{2}$

Substitute the value of each term in the equation to calculate the height reached by the balloon.

$S=0\times 5+\dfrac{1}{2}\times 10\times 5^{2}$

$S= 125 m$

Now, even after the stone is released, it will continue to move upward for some time because it has some velocity in the upward direction. Then, the initial velocity of the stone is equal to the final velocity of the balloon after 5 sec. So, let us calculate the final velocity of the balloon using the formula given below:

$v=u+at$

Substitute the values and calculate the final velocity.

$v=0+10\times 5$

$v= 50 m/s$

Now, calculate the distance covered by the stone to reach the maximum height where the final velocity is zero by using the formula given below. (Take acceleration due to gravity as -10 m/s2 since it is in the downward direction.)

$v^2=u^2+2as$

$0^2=50^2+2\times (-10)\times$

$20s=2500$

$s=125 m$

Therefore, the maximum height attained by the stone is the sum of the distance travelled by the balloon in 5 sec and the distance travelled by the stone after it is released, i.e, 125 m + 125 m = 250 m

Key Point: The velocity of any body thrown upward will be zero when it reaches its highest point but the acceleration will be equal to the acceleration due to gravity acting downwards.

2. An aeroplane moving at a speed of 180 km/hr drops a food packet while flying at a height of 500 m. Find the horizontal range?

Ans: We will consider the motion along with vertical and horizontal directions separately to apply equations of motion.

First, find the time taken by the packet to reach the ground considering motion along the vertical direction.

The vertical displacement of the packet, Sy = h = 500 m

Initial velocity along vertical direction, uy = 0 m/s

Acceleration of the packet along vertical direction = g = 10 m/s2

$s_y=u_yt+\dfrac{1}{2}a_yt^{2}$

$S_y=u_yt+\dfrac{1}{2}a_yt^{2}$

$500 =0\times t+\dfrac{1}{2}\times 10t^{2}$

$1000=10t^{2}$

$t^2=100$

$t=10s$

Now consider the motion along horizontal direction and calculate the horizontal range using the equation of motion.

Initial velocity along the horizontal direction, ux = 180 km/hr = $180\times \dfrac{5}{18}$ = 50 m/s

Acceleration along the horizontal direction, ax = 0 m/s2

Time taken to reach the ground, t = 10s

$S_x=u_xt+\dfrac{1}{2}a_xt^{2}$

$S=50\times10+\dfrac{1}{2}\times 0\times 10^{2}$

$S_x= 500 m$

Key Point: Whenever dealing with the problems related to the two-dimensional motion of a body under gravity, consider the motion along with vertical and horizontal directions separately to apply equations of motion.

1. Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by bullets fired by two guns on the ground is… (JEE 2019)

Ans: For a projectile motion, maximum range is attained at an angle of 45o. The area covered by the bullets will be equal to the area of the circle with a radius equal to the maximum range of each bullet.

Bullets fired by gun A:

Let RA be the maximum range of the bullets fired by gun A at an angle of 45o.

The velocity of the bullets fired by gun A, uA = 1 km/s

The maximum range of the bullet from gun A is given by

$R_A=\dfrac{u_A^2\sin(2\theta)}{g}$

$R_A=\dfrac{u_A^2\sin(2\times45)}{g}$

$R_A=\dfrac{u_A^2}{g}$

Maximum area covered by the bullets from gun A is

$Area_A=\pi R_A^2$

$Area_A=\pi\left( \dfrac{u_A^4}{g^2}\right) $

Similarly,

The velocity of the bullets fired by gun B, uB = 2 km/s

The maximum range of the bullet from gun B is given by

$R_B=\dfrac{u_B^2}{g}$

Maximum area covered by the bullets from gun B is

$Area_B=\pi R_B^2$

$Area_B=\pi\left( \dfrac{u_B^4}{g^2}\right) $

Then the ratio of two area covered by the bullets fired by gun A and gun B is,

$\dfrac{Area_A}{Area_B}=\dfrac{\pi\left( \dfrac{u_A^4}{g^2}\right)}{\pi\left( \dfrac{u_B^4}{g^2}\right)}$

$\dfrac{Area_A}{Area_B}=\dfrac{u_A^4}{u_B^4}=\dfrac{1^4}{2^4}=\dfrac{1}{16} $ Therefore, the ratio of area covered by the bullets A and B is 1:16.

Key Point: The maximum range is attained by a given projectile for a given velocity at an angle of 450.

2. A stone is dropped from the top of a building. When it crosses a point 5m below the top, another stone starts to fall from a point 25m below the top, both stones reach the bottom of the building simultaneously. The height of the building is: (Take g = 10 m/s^{2}) (JEE 2021 Feb)

Ans: First, we have to calculate the velocity of the first stone (u1) at a height 5 m below the top of the building using the formula given below.

$u_1=\sqrt {2gh}$

$u_1=\sqrt {2\times 10\times 5}=10m/s$

Now, apply the 2nd equation of motion on the first stone at a point 5 m below from the top till it reaches the ground. Then the displacement covered by the first stone is 20+h.

$20+h=u_1t+\dfrac{1}{2}gt^{2}$

$20+h=10t+\dfrac{1}{2}gt^{2}$.....(1)

For the second stone, the displacement covered to reach the ground is h and the time taken is t. Using the 2nd equation of motion,

$20+h=10t+\dfrac{1}{2}gt^{2}$.....(2)

Put equation (2) in equation (1) to calculate the time t

$20+\dfrac{1}{2}gt^{2}=10t+\dfrac{1}{2}gt^{2}$

t=2s

Now, put t=2s in equation (2) to calculate h

$h=\dfrac{1}{2}\times 10\times 2^{2}=20 m$

The total height of the building = h + 25= 20 + 25 = 45 m

Key Point: Whenever an object is dropped or released, the initial velocity must be taken as zero.

A body is projected at 600 with the ground. It covers a horizontal distance of 100 m. If the same body is projected at 600 with vertical with the same velocity, what will be the new range? (Ans: 100 m)

A car moving with a speed of 50 km/hr, can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100 km/hr, what will be the minimum stopping distance? (Ans: 24 m)

So here we discussed important concepts and formulas related to **Kinematics** from the JEE point of view. Students must make sure that they do not miss any of the above important topics to obtain a good score in the **JEE Main exam**.

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Last updated date: 19th Mar 2023

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NTA has announced the JEE Main 2023 January session application form release date on the official website https://jeemain.nta.nic.in/. JEE Main 2023 January and April session Application Form is available on the official website for online registration. Besides JEE Main 2023 January and April session application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details. April Session's details will be updated soon by NTA.

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It is crucial for the the engineering aspirants to know and download the JEE Main 2023 syllabus PDF for Maths, Physics and Chemistry. Check JEE Main 2023 syllabus here along with the best books and strategies to prepare for the entrance exam. Download the JEE Main 2023 syllabus consolidated as per the latest NTA guidelines from Vedantu for free.

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In order to prepare for JEE Main 2023, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2023 exam so that they can grab the top rank in the all India entrance exam.

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JEE Main 2023 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2023 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.

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NTA is responsible for the release of the JEE Main 2023 January and April Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2023 January and April Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.

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NTA will release the JEE Main 2023 January and April sessions exam dates on the official website, i.e. {official-website}. Candidates can directly check the date sheet on the official website or https://jeemain.nta.nic.in/. JEE Main 2023 January and April sessions is expected to be held in February and May. Visit our website to keep updates of the respective important events of the national entrance exam.

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Want to know which Engineering colleges in India accept the JEE Main 2023 scores for admission to Engineering? Find the list of Engineering colleges accepting JEE Main scores in India, compiled by Vedantu. There are 1622 Colleges that are accepting JEE Main. Also find more details on Fees, Ranking, Admission, and Placement.

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FAQ

**1. How many questions are asked from Kinematics in JEE Main?**

Mostly, 1-2 questions from this chapter are asked in JEE Main every year which corresponds to around 8 marks in the JEE Main exam.

**2. What is the importance of kinematics in the JEE exam?**

Compared to other chapters, kinematics concepts will become easier for you once you start practicing more and more questions related to it. You can easily solve all the types of kinematics problems asked from this chapter once you learn all the topics covered in the chapter conceptually. Besides, the concepts discussed in this chapter are linked to problems from other chapters too.

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