Prepare effectively for JEE Main with Topicwise PYQs, Previous Year Questions, MCQs, and detailed solutions from Motion in a Straight Line. This collection covers all important concepts, including distance and displacement, speed and velocity, acceleration, equations of motion, graphical analysis of motion, and relative motion. Practicing these carefully selected questions helps students strengthen conceptual understanding, improve problem-solving skills, and gain confidence for JEE Main, Class 11 Physics, and other competitive examinations.
Motion in a Straight Line IIT JEE Main PYQs, MCQs and Topicwise Most Important Questions with Solutions
Master the chapter Motion in a Straight Line with JEE Main previous year questions, important MCQs, and topicwise practice papers. These questions help students develop a strong understanding of fundamental concepts and improve their ability to solve numerical and conceptual problems efficiently.
Gain deeper understanding by studying JEE Main Motion in a Straight Line MCQs PYQs Previous Year Questions and Solutions
Why is regular question practice important for JEE Main Physics?
Regular practice helps students improve accuracy, increase problem-solving speed, and develop confidence in handling different types of examination questions. It also strengthens conceptual understanding and reduces mistakes during the actual exam.
“Students should also study Equations of Motion Under Gravity for free fall, upward and vertical downward motion“
How can topicwise questions help in exam preparation?
Topicwise questions allow students to focus on one concept at a time, identify weak areas, and build a solid foundation before moving to advanced problems. This approach makes revision more organized and effective.
“Read more about Derive Equations of Motion Using Calculus Method, Solved Numerical Examples“
JEE Main Question
A ball is thrown up with a certain velocity, so that it reaches a height h. Find the ratio of the two different times of the ball reaching h/3 in both directions.
(A) $\frac{\sqrt{2} – 1}{\sqrt{2} + 1}$
(B) $\frac{1}{3}$
(C) $\frac{\sqrt{3} – \sqrt{2}}{\sqrt{3} + \sqrt{2}}$
(D) $\frac{\sqrt{3} – 1}{\sqrt{3} + 1}$
(JEE Main 2021, 27 July Shift-I)
Solution
Answer: (C)
When a ball is thrown vertically upward with an initial velocity $u$, it experiences a downward acceleration due to gravity (a = –g). At its maximum height h, its final velocity $v = 0$.
Using the third equation of motion ($v^2 = u^2 + 2as$) :
$$0 = u^2 + 2(-g)h$$
$$u^2 = 2gh \implies u = \sqrt{2gh} $$
Using the second equation of motion ($s = ut + \frac{1}{2}at^2$), we substitute the displacement $s = \frac{h}{3}$ and acceleration a = –g :
$$\frac{h}{3} = ut – \frac{1}{2}gt^2$$
$$\frac{1}{2}gt^2 – ut + \frac{h}{3} = 0$$
$$3gt^2 – 6ut + 2h = 0 $$
Substitute the value of $u = \sqrt{2gh}$ :
$$3gt^2 – 6\left(\sqrt{2gh}\right)t + 2h = 0$$
$$t = \frac{-(-6\sqrt{2gh}) \pm \sqrt{(-6\sqrt{2gh})^2 – 4(3g)(2h)}}{2(3g)}$$
$$t = \frac{6\sqrt{2gh} \pm \sqrt{36(2gh) – 24gh}}{6g}$$
$$t = \frac{6\sqrt{2gh} \pm \sqrt{72gh – 24gh}}{6g}$$
$$t = \frac{6\sqrt{2gh} \pm \sqrt{48gh}}{6g}$$
$$t = \frac{6\sqrt{2}\sqrt{gh} \pm 4\sqrt{3}\sqrt{gh}}{6g}$$
$$t = \frac{2\sqrt{gh} \left( 3\sqrt{2} \pm 2\sqrt{3} \right)}{6g}$$
$$t = \frac{\sqrt{gh}}{3g} \left( 3\sqrt{2} \pm 2\sqrt{3} \right)$$
This gives us two distinct times, $t_1$ (upward journey, smaller time) and $t_2$ (downward journey, larger time) :
$t_1 = \dfrac{\sqrt{gh}}{3g} \left( 3\sqrt{2} – 2\sqrt{3} \right)$
$t_2 = \dfrac{\sqrt{gh}}{3g} \left( 3\sqrt{2} + 2\sqrt{3} \right)$
Divide $t_1$ by $t_2$ :
$$\frac{t_1}{t_2} = \frac{3\sqrt{2} – 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}}$$
$$\frac{t_1}{t_2} = \frac{\sqrt{6}(\sqrt{3} – \sqrt{2})}{\sqrt{6}(\sqrt{3} + \sqrt{2})}$$
$$\frac{t_1}{t_2} = \frac{\sqrt{3} – \sqrt{2}}{\sqrt{3} + \sqrt{2}}$$
The ratio of the two different times of the ball reaching $\dfrac{h}{3}$ is $\dfrac{\sqrt{3} – \sqrt{2}}{\sqrt{3} + \sqrt{2}}$.
Hence, the correct option is (C).
“Build strong concepts by studying Derive Equations of Uniformly Accelerated Motion“
JEE Main Question
A balloon was moving upwards with a uniform velocity of 10 m/s. An object of finite mass is dropped from the balloon when it was at a height of 75 m from the ground level. The height of the balloon from the ground when the object strikes the ground was around (Take g = 10 m/s2) :
(A) 300 m
(B) 200 m
(C) 125 m
(D) 250 m
(JEE Main 2021, 25 July Shift-II)
Solution
Answer: (C)
When an object is dropped from a moving carrier (like a rising balloon), it inherits the instantaneous velocity of that carrier due to inertia. Therefore, the object initially moves upward at 10 m/s relative to the ground before turning around and falling down. Let the point where the object is dropped be the origin.
Using the sign convention : Upward direction = Positive (+), Downward direction = Negative (-)
For the dropped object :
Initial velocity, $u$ = +10 m/s (upward)
Acceleration, a = –g = -10 m/s2 (downward)
Net displacement when it hits the ground, s = -75 m (since the ground is 75 m below the drop point)
Using the second equation of motion ($s = ut + \frac{1}{2}at^2$) :
$$-75 = 10t + \frac{1}{2}(-10)t^2$$
$$-75 = 10t – 5t^2$$
$$5t^2 – 10t – 75 = 0$$
$$t^2 – 2t – 15 = 0$$
$$t^2 – 5t + 3t – 15 = 0$$
$$t(t – 5) + 3(t – 5) = 0$$
$$(t – 5)(t + 3) = 0$$
This gives two roots : $t = 5\text{ s}$ or $t = -3\text{ s}$. Since time cannot be negative, we take :
$$t = 5 \text{ s}$$
The balloon continues to move upward at a constant, uniform velocity of $v_b = 10 \text{ m/s}$ for the entire duration of the object’s flight ($5\text{ s}$).
$$\text{Distance moved by balloon } (s_b) = \text{velocity} \times \text{time}$$
$$s_b = 10 \text{ m/s} \times 5 \text{ s} = 50 \text{ m}$$
The total height of the balloon from the ground when the object strikes the floor is the sum of its initial height and the extra distance it climbed :
$$\text{Final Height} = \text{Initial Height} + s_b$$
$$\text{Final Height} = 75 \text{ m} + 50 \text{ m} = 125 \text{ m}$$
The height of the balloon from the ground when the object strikes the ground is 125 m.
Hence, the correct option is (C).
“For complete preparation, also study JEE Units and Measurements PYQs Previous Year Questions and Solutions“
JEE Main Question
Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at the 4th second after its fall to the next droplet is 34.3 m. At what rate are the droplets coming from the tap? (Take g = 9.8 m/s2)
(A) 3 drops / 2 s
(B) 2 drops / s
(C) 1 drop / s
(D) 1 drop / 7 s
(JEE Main 2021, 25 July Shift-I)
Solution
Answer: (C)
The first droplet has been falling freely from rest ($u = 0$) for exactly $t_1 = 4\text{ s}$. Using the second equation of motion ($s = ut + \frac{1}{2}gt^2$) :
$$s_1 = \frac{1}{2}gt_1^2$$
Substitute $g = 9.8\text{ m/s}^2$ and $t_1 = 4\text{ s}$ :
$$s_1 = \frac{1}{2} \times 9.8 \times 4^2$$
$$s_1 = 4.9 \times 16 = 78.4\text{ m}$$
We are given that at this instant, the separation between this first droplet and the next consecutive (second) droplet is $34.3\text{ m}$.
Therefore, the distance fallen by the second droplet from the tap is :
$$s_2 = s_1 – \text{separation}$$
$$s_2 = 78.4\text{ m} – 34.3\text{ m} = 44.1\text{ m}$$
Using the same equation of motion for the second droplet to find how long it has been in the air :
$$s_2 = \frac{1}{2}gt_2^2$$
$$44.1 = \frac{1}{2} \times 9.8 \times t_2^2$$
$$44.1 = 4.9 \times t_2^2$$
$$t_2^2 = \frac{44.1}{4.9} = \frac{441}{49} = 9$$
$$t_2 = 3\text{ s}$$
The first droplet has been falling for $4\text{ s}$.
The second droplet has been falling for $3\text{ s}$.
The time interval ($\Delta t$) between the release of the first and second droplet is :
$$\Delta t = t_1 – t_2 = 4\text{ s} – 3\text{ s} = 1\text{ s}$$
Since the droplets fall at a constant, regular rate, a new droplet leaves the tap every 1 second.
The rate at which droplets are coming from the tap is 1 drop / s.
Hence, the correct option is (C).
For complete preparation, also study NCERT Exemplar Solutions Motion In a Straight Line Class 11 Physics Chapter 2
JEE Main Question
A scooter accelerates from rest for time $t_1$ at a constant rate $\alpha_1$ and then retards at a constant rate $\alpha_2$ for time $t_2$ and comes to rest. The correct value of $\dfrac{t_1}{t_2}$ will be :
(A) $\frac{\alpha_1 + \alpha_2}{\alpha_2}$
(B) $\frac{\alpha_2}{\alpha_1}$
(C) $\frac{\alpha_1}{\alpha_2}$
(D) $\frac{\alpha_1 + \alpha_2}{\alpha_1}$
(JEE Main 2021, 26 Feb Shift-II)
Solution
Answer: (B)
The scooter starts from rest, so its initial velocity is $u = 0$. It accelerates at a constant rate $\alpha_1$ for a time duration $t_1$ until it reaches a maximum velocity, let’s call it $v_{\text{max}}$.
Using the first equation of motion ($v = u + at$) :
$$v_{\text{max}} = 0 + \alpha_1 t_1$$
$$v_{\text{max}} = \alpha_1 t_1 $$
The scooter then begins to decelerate from its maximum velocity $v_{\text{max}}$ at a constant rate $\alpha_2$ for a time duration $t_2$ until it finally comes back to rest ($v_{\text{final}} = 0$).
Using the first equation of motion ($v = u – at$ for deceleration) :
$$0 = v_{\text{max}} – \alpha_2 t_2$$
$$v_{\text{max}} = \alpha_2 t_2 $$
Since the maximum velocity $v_{\text{max}}$ reached at the end of the acceleration phase is the exact same velocity from which the retardation phase begins, we can equate above equations :
$$\alpha_1 t_1 = \alpha_2 t_2$$
$$\frac{t_1}{t_2} = \frac{\alpha_2}{\alpha_1}$$
The correct value of the ratio $\dfrac{t_1}{t_2}$ is $\dfrac{\alpha_2}{\alpha_1}$.
Hence, the correct option is (B).
To strengthen your concepts, learn about Relative Velocity Formula Explained With Solved Examples
JEE Main Question
A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of the building simultaneously. The height of the building is :
(a) 45 m
(b) 25 m
(c) 35 m
(d) 50 m
[JEE Main 2021, 25 Feb Shift-II]
Solution
Answer: (a)
![[2021, 25 Feb Shift-II] Solved JEE Main PYQ Numerical Problem](https://physicsanandclasses.co.in/wp-content/uploads/2026/05/2021-25-Feb-Shift-II-Solved-JEE-Main-PYQ-Problem.webp "JEE Main PYQs Previous Year Questions MCQs Topicwise Papers and Solutions Motion in a Straight Line 2")
Let the top of the building be T and the ground be G. The total height of the building is x. Both stones are dropped from rest, so their initial velocities at their respective release points are zero.
The first stone falls a distance of 5 m from the top to reach point A.
Using the equation of motion $s = ut + \frac{1}{2}gt^2$ with $u = 0$ and $g = 10\text{ m/s}^2$ :
$$5 = 0 + \frac{1}{2}(10)t_1^2$$
$$5 = 5t_1^2 \implies t_1 = 1\text{ s}$$
Now, find its velocity ($v_A$) as it passes point A using $v = u + gt$ :
$$v_A = 0 + (10)(1) = 10\text{ m/s}$$
Let t be the remaining time it takes for both stones to hit the ground simultaneously after the second stone is released.
For the second stone (from B to G) : It falls from rest from a point 25 m below the top. The remaining distance to the ground is x – 25.$$x – 25 = 0 + \frac{1}{2}gt^2$$$$x – 25 = 5t^2 $$
For the first stone (from A to G) : It travels from point A to the ground. The remaining distance is x – 5, and its speed at A is $v_A = 10\text{ m/s}$.$$x – 5 = v_A t + \frac{1}{2}gt^2$$$$x – 5 = 10t + 5t^2 $$
Substitute the value of 5t2 :
x – 5 = 10t + (x – 25)
-5 = 10t – 25
t = 2 s
Now substitute t = 2 s to find x :
x – 25 = 5(2)2
x – 25 = 20
x = 45 m
The total height of the building is 45 m.
Hence, the correct option is (a).
JEE Main Question
An engine of a train moving with uniform acceleration passes a signal post with velocity $u$ and the last compartment passes it with velocity $v$. The velocity with which the middle point of the train passes the signal post is :
(a) $\sqrt{\frac{v^2 + u^2}{2}}$
(b) $\frac{v – u}{2}$
(c) $\frac{u + v}{2}$
(d) $\sqrt{\frac{v^2 – u^2}{2}}$
[JEE Main 2021, 25 Feb Shift-I]
Solution
Answer: (a)
Let the total length of the train be L. The train is moving with a constant acceleration a.
We can break the motion down into two parts relative to the stationary signal post:
Step 1: Motion of the entire train
When the front of the train (engine) passes the post, its speed is $u$. When the back of the train (last compartment) passes the post, the train has traveled a distance equal to its entire length L, and its speed becomes $v$.
Using the third equation of motion ($v^2 = u^2 + 2as$) :
$$v^2 = u^2 + 2aL$$
$$2aL = v^2 – u^2 \implies aL = \frac{v^2 – u^2}{2} $$
Step 2: Motion up to the middle point
Let $v’$ be the velocity of the train when its middle point passes the signal post. At this exact moment, the front of the train has traveled a distance equal to half of the train’s total length ($s = \frac{L}{2}$).
Using the third equation of motion again, with initial velocity $u$ and final velocity $v’$ :
$$(v’)^2 = u^2 + 2a\left(\frac{L}{2}\right)$$
$$(v’)^2 = u^2 + aL $$
Step 3: Calculate the middle-point velocity ($v’$)
Substitute the value of $aL$ :
$$(v’)^2 = u^2 + \left(\frac{v^2 – u^2}{2}\right)$$
$$(v’)^2 = \frac{2u^2 + v^2 – u^2}{2}$$
$$(v’)^2 = \frac{v^2 + u^2}{2}$$
$$v’ = \sqrt{\frac{v^2 + u^2}{2}}$$
The velocity with which the middle point of the train passes the signal post is $\sqrt{\dfrac{v^2 + u^2}{2}}$.
Hence, the correct option is (a).
JEE Main Question
A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to (Here, g is the acceleration due to gravity) :
(a) $t = \sqrt{\frac{2h}{3g}}$
(b) $t = 1.8\sqrt{\frac{h}{g}}$
(c) $t = 2\sqrt{\frac{h}{3g}}$
(d) $t = 3.4\sqrt{\frac{h}{g}}$
(JEE Main 2020, 5 Sep Shift-I)
Solution
Answer: (d)
The helicopter starts from rest ($u_h = 0$) on the ground and moves vertically upwards with a constant acceleration $a = +g$.
Using the third equation of motion ($v^2 = u^2 + 2as$) to find its velocity $v$ at height $h$ :
$$v^2 = 0 + 2(g)(h)$$
$$v = \sqrt{2gh}$$
When the food packet is released, it inherits this upward velocity of the helicopter due to inertia.
Let’s analyze the motion of the food packet from the moment it is dropped until it hits the ground.
Sign convention : Upward is positive (+), downward is negative (-).
Initial velocity of packet, $u = +\sqrt{2gh}$ (upward)
Acceleration on packet, $a = -g$ (downward gravity)
Net displacement of packet, $s = -h$ (since the ground is a distance $h$ below the release point)
Using the second equation of motion ($s = ut + \frac{1}{2}at^2$) :
$$-h = \left(\sqrt{2gh}\right)t + \frac{1}{2}(-g)t^2$$
$$\frac{1}{2}gt^2 – \left(\sqrt{2gh}\right)t – h = 0$$
$$gt^2 – 2\left(\sqrt{2gh}\right)t – 2h = 0$$
$$t = \frac{-\left(-2\sqrt{2gh}\right) \pm \sqrt{\left(-2\sqrt{2gh}\right)^2 – 4(g)(-2h)}}{2g}$$
$$t = \frac{2\sqrt{2gh} \pm \sqrt{4(2gh) + 8gh}}{2g}$$
$$t = \frac{2\sqrt{2gh} \pm \sqrt{8gh + 8gh}}{2g}$$
$$t = \frac{2\sqrt{2gh} \pm \sqrt{16gh}}{2g}$$
Since $\sqrt{16gh} = 4\sqrt{gh}$:
$$t = \frac{2\sqrt{2}\sqrt{gh} \pm 4\sqrt{gh}}{2g}$$
We ignore the negative root because time must be positive:
$$t = \frac{\sqrt{gh}}{2g} \left(2\sqrt{2} + 4\right)$$
$$t = \frac{\sqrt{gh}}{g} \left(\sqrt{2} + 2\right)$$
Since $\frac{\sqrt{gh}}{g} = \sqrt{\frac{h}{g}}$:
$$t = \left(\sqrt{2} + 2\right)\sqrt{\frac{h}{g}}$$
Substitute the value of $\sqrt{2} \approx 1.414$ :
$$t = (1.414 + 2)\sqrt{\frac{h}{g}}$$
$$t = 3.414\sqrt{\frac{h}{g}} \approx 3.4\sqrt{\frac{h}{g}}$$
The time taken by the packet to reach the ground is close to $3.4\sqrt{\dfrac{h}{g}}$.
Hence, the correct option is (d).
JEE Main Question
From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground is n times that taken by it to reach the highest point of its path. The relation between H, u and n is :
(a) $2gH = n^2u^2$
(b) $gH = (n – 2)^2u^2$
(c) $2gH = nu^2(n – 2)$
(d) $gH = (n – 2)u^2$
(JEE Main 2014)
Solution
Answer: (c)
Let time taken to reach the highest point is ($t_1$). When a particle is thrown vertically upwards with an initial velocity $u$, its final velocity at the highest point is $v = 0$. The acceleration is due to gravity acting downwards ($a = -g$).
Using the first equation of motion ($v = u + at$) :
$$0 = u – gt_1$$
$$t_1 = \frac{u}{g}$$
Let the total time to hit the ground is ($t_2$). According to the given condition, the total time ($t_2$) taken by the particle to reach the ground is $n$ times the time taken to reach the highest point ($t_1$) :
$$t_2 = n \cdot t_1 = \frac{nu}{g}$$
Let’s analyze the net displacement from the top of the tower to the ground.
Sign convention : Upward is positive (+), downward is negative (-).
Initial velocity, $u = +u$ (upward)
Acceleration, $a = -g$ (downward)
Total time of flight, $t = t_2 = \dfrac{nu}{g}$
Net displacement, $s = -H$ (since the ground is a distance $H$ below the projection point)
Using the second equation of motion ($s = ut + \frac{1}{2}at^2$) :
$$-H = u(t_2) – \frac{1}{2}g(t_2)^2$$
Substitute the value of $t_2$ :
$$-H = u\left(\frac{nu}{g}\right) – \frac{1}{2}g\left(\frac{nu}{g}\right)^2$$
$$-H = \frac{nu^2}{g} – \frac{1}{2}g\left(\frac{n^2u^2}{g^2}\right)$$
$$-H = \frac{nu^2}{g} – \frac{n^2u^2}{2g}$$
$$2gH = -2nu^2 + n^2u^2$$
$$2gH = n^2u^2 – 2nu^2$$
$$2gH = nu^2(n – 2)$$
The correct relation between $H$, $u$, and $n$ is $2gH = nu^2(n – 2)$.
Hence, the correct option is (c).
JEE Main Question
A parachutist after bailing out falls 50 m without friction. When the parachute opens, it decelerates at 2 ms-2. He reaches the ground with a speed of 3 m/s. At what height did he bail out?
(a) 91 m
(b) 182 m
(c) 293 m
(d) 111 m
(AIEEE 2005)
Solution
Answer: (c)
Let the total height from which the parachutist bails out be H. The motion is divided into two parts : free fall under gravity, followed by controlled deceleration after opening the parachute.
Part 1: Free fall before the parachute opens
The parachutist jumps from rest ($u_1 = 0$) and falls a distance $h = 50\text{ m}$ under gravity ($g = 9.8\text{ m/s}^2$) without friction.
Using the third equation of motion ($v^2 = u^2 + 2as$) to find his velocity ($v$) right before the parachute opens:
$$v^2 = 0 + 2gh$$
$$v^2 = 2 \times 9.8 \times 50$$
$$v^2 = 980\text{ (m/s)}^2$$
Part 2: Deceleration after the parachute opens
The velocity $v$ acts as the initial velocity for this phase. The parachute opens, causing a uniform retardation (negative acceleration) until the parachutist reaches the ground.
Initial velocity for this phase, $u_2^2 = v^2 = 980\text{ (m/s)}^2$
Final velocity at the ground, $v_2 = 3\text{ m/s} \implies v_2^2 = 9\text{ (m/s)}^2$
Deceleration, $a = -2\text{ m/s}^2$
Distance traveled during deceleration = $y$
Using the third equation of motion again ($v_2^2 = u_2^2 + 2ay$) :
$$9 = 980 + 2(-2)y$$
$$9 = 980 – 4y$$
$$4y = 980 – 9$$
$$4y = 971$$
$$y = \frac{971}{4} = 242.75\text{ m}$$
The total height from which he bailed out is the sum of the distance fallen during free fall ($h$) and the distance traveled during deceleration ($y$) :
$$H = h + y$$
$$H = 50\text{ m} + 242.75\text{ m} = 292.75\text{ m}$$
The parachutist bailed out at a height of 293 m.
Hence, the correct option is (c).
JEE Main Question
A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?
(a) h/9 m from the ground
(b) 7h/9 m from the ground
(c) 8h/9 m from the ground
(d) 17h/18 m from the ground
(AIEEE 2004)
Solution
Answer: (c)
The ball is dropped from rest, so its initial velocity is $u = 0$. It falls for a total time $T$ to cover the total height $h$.
Using the second equation of motion ($s = ut + \frac{1}{2}gt^2$) :
$$h = 0 + \frac{1}{2}gT^2 \implies h = \frac{1}{2}gT^2 $$
Let $y$ be the downward distance traveled by the ball from the top of the tower in the first $t = \frac{T}{3}$ seconds.
Using the equation of motion ($s = ut + \frac{1}{2}gt^2$) :
$$y = 0 + \frac{1}{2}g\left(\frac{T}{3}\right)^2$$
$$y = \frac{1}{2}g\left(\frac{T^2}{9}\right)$$
$$y = \frac{1}{9} \left(\frac{1}{2}gT^2\right)$$
Since $\frac{1}{2}gT^2 = h$ :
$$y = \frac{h}{9}$$
So, the ball has fallen a distance of $\frac{h}{9}$ downwards from the top.
The question specifically asks for the position measured from the ground.
$$\text{Height from ground} = \text{Total height} – \text{Distance fallen}$$
$$\text{Height from ground} = h – \frac{h}{9}$$
$$\text{Height from ground} = \frac{9h – h}{9} = \frac{8h}{9}$$
The position of the ball after $T/3$ seconds is $8h/9\text{ m}$ from the ground.
Hence, the correct option is (c).
JEE Main Question
An automobile travelling with a speed of 60 km/h can brake to stop within a distance of 20 m. If the car is going twice as fast, i.e., 120 km/h, the stopping distance will be :
(a) 20 m
(b) 40 m
(c) 60 m
(d) 80 m
(AIEEE 2004)
Solution
Answer: (d)
When an automobile brakes to a halt, its final velocity becomes zero ($v = 0$). Assuming the braking mechanism provides the same constant retardation ($a$) in both situations, we can analyze the stopping distance using the equations of motion.
Using the third equation of motion ($v^2 = u^2 + 2as$), where $v = 0$ and acceleration is negative due to braking (retardation):
$$0 = u^2 – 2as$$
$$2as = u^2 \implies s = \frac{u^2}{2a}$$
Since the braking retardation ($a$) is constant, the stopping distance is directly proportional to the square of the initial speed :
$$s \propto u^2$$
Let the initial speed in the first case be $u_1 = 60\text{ km/h}$ and the corresponding stopping distance be $s_1 = 20\text{ m}$.
In the second case, the initial speed is doubled:
$$u_2 = 120\text{ km/h} = 2u_1$$
Setting up the ratio for both cases :
$$\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2$$
Substitute $u_2 = 2u_1$ into the ratio :
$$\frac{s_2}{s_1} = \left(\frac{2u_1}{u_1}\right)^2 = (2)^2 = 4$$
Now, solve for the new stopping distance ($s_2$) :
$$s_2 = 4 \times s_1$$
$$s_2 = 4 \times 20\text{ m} = 80\text{ m}$$
If the car travels twice as fast, its stopping distance increases by a factor of 4, making it 80 m.
Hence, the correct option is (d).
JEE Main Question
A car moving with a speed of 50 km/h can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/h, the minimum stopping distance is :
(a) 12 m
(b) 18 m
(c) 24 m
(d) 6 m
(AIEEE 2003)
Solution
Answer: (c)
When a car is brought to a stop by applying brakes, its final velocity becomes zero ($v = 0$). Assuming the brakes provide the same maximum constant deceleration (retardation, $a$) in both situations, we can determine the new stopping distance.
From the third equation of motion ($v^2 = u^2 + 2as$), since the car comes to rest ($v = 0$) and undergoes retardation :
$$0 = u^2 – 2as$$
Rearranging the terms to express stopping distance ($s$) :
$$2as = u^2 \implies s = \frac{u^2}{2a}$$
Because the braking deceleration ($a$) depends on the car’s brakes and tires (which remain the same), $a$ is a constant. Therefore, the stopping distance is directly proportional to the square of the initial speed :
$$s \propto u^2$$
Let the initial speed in the first scenario be $u_1 = 50\text{ km/h}$ and the corresponding minimum stopping distance be $s_1 = 6\text{ m}$.
In the second case, the speed is doubled :
$$u_2 = 100\text{ km/h} = 2u_1$$
Setting up a ratio between the two cases :
$$\frac{s_2}{s_1} = \left(\frac{u_2}{u_1}\right)^2$$
Substitute $u_2 = 2u_1$ into the relation :
$$\frac{s_2}{s_1} = \left(\frac{2u_1}{u_1}\right)^2 = (2)^2 = 4$$
Now, calculate the new minimum stopping distance ($s_2$) :
$$s_2 = 4 \times s_1$$
$$s_2 = 4 \times 6\text{ m} = 24\text{ m}$$
When the speed of the car is doubled, its minimum stopping distance increases by a factor of 4, resulting in 24 m.
Hence, the correct option is (c).
JEE Main Question
The speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distances at which the two cars are stopped from that instant is :
(a) 1:1
(b) 1:4
(c) 1:8
(d) 1:16
(AIEEE 2002)
Solution
Answer: (d)
The problem states that the two cars are identical. This implies that the maximum braking force they can apply and the coefficient of friction ($\mu$) between their tires and the road surface are exactly the same. As a result, both cars will undergo the same constant deceleration (retardation, a).
When a car is brought to a stop, its final velocity becomes zero ($v = 0$). Using the third equation of motion ($v^2 = u^2 + 2as$), where the acceleration is negative due to retardation :
$$0 = u^2 – 2as$$
$$2as = u^2 \implies s = \frac{u^2}{2a}$$
Since the deceleration (a) is identical for both cars, the stopping distance is directly proportional to the square of the initial speed :
$$s \propto u^2$$
Let $s_1$ be the stopping distance for the first car with initial speed $u_1 = u$, and let $s_2$ be the stopping distance for the second car with initial speed $u_2 = 4u$.
$$\frac{s_1}{s_2} = \left(\frac{u_1}{u_2}\right)^2$$
Substitute the given initial speeds into the ratio :
$$\frac{s_1}{s_2} = \left(\frac{u}{4u}\right)^2$$
$$\frac{s_1}{s_2} = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$$
$$s_1 : s_2 = 1 : 16$$
The ratio of the respective distances at which the two cars are stopped is $1:16$.
Hence, the correct option is (d).
JEE Main Question
If a body loses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest?
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm
(AIEEE 2002)
Solution
Answer: (a)
Assume the body undergoes a uniform resistance (constant deceleration, $a$) as it penetrates the wooden block.
We can divide the motion of the body into two consecutive parts :
Part 1: Initial penetration of 3 cm
Initial velocity, $u_1 = v$
Final velocity, $v_1 = \frac{v}{2}$ (since it loses half of its velocity)
Distance traveled, $s_1 = 3\text{ cm}$
Using the third equation of motion ($v^2 = u^2 + 2as$) :
$$\left(\frac{v}{2}\right)^2 = v^2 + 2a(3)$$
$$\frac{v^2}{4} = v^2 + 6a$$
$$6a = \frac{v^2}{4} – v^2$$
$$6a = -\frac{3v^2}{4}$$
$$a = -\frac{3v^2}{24} = -\frac{v^2}{8} $$
(The negative sign confirms that it is deceleration).
Part 2: Motion from that point until it comes to rest
Let $x$ be the additional distance the body penetrates before coming to a complete stop.
Initial velocity for this phase, $u_2 = \frac{v}{2}$
Final velocity, $v_2 = 0$ (comes to rest)
Distance traveled, $s_2 = x$
Using the third equation of motion again ($v_2^2 = u_2^2 + 2as_2$) :
$$0 = \left(\frac{v}{2}\right)^2 + 2ax$$
$$0 = \frac{v^2}{4} + 2ax$$
$$2ax = -\frac{v^2}{4} $$
Substitute the value of a :
$$2\left(-\frac{v^2}{8}\right)x = -\frac{v^2}{4}$$
$$-\frac{v^2}{4} \cdot x = -\frac{v^2}{4}$$
$$x = 1\text{ cm}$$
The body will penetrate 1 cm more before coming to rest.
Hence, the correct option is (a).
JEE Main PYQs Previous Year Questions With Solutions of Chapter Units and Measurements Class 11 Physics
Build a strong foundation by first exploring Class 11 Physics Notes for clear concepts and complete theory.
Next, understand the basics in depth through the Units and Measurements chapter to strengthen your fundamentals.
Once your concepts are clear, practice Units and Measurements JEE Main PYQs Set-1 to get familiar with question patterns.
Then move to Units and Measurements JEE Main PYQs Set-2 to improve accuracy and problem-solving speed.
Further enhance your preparation with Units and Measurements JEE Main PYQs Set-3 for advanced practice and better exam readiness.