JEE Main Motion in Straight Line PYQ

6. (JEE Main 2024 (Online) 30th January Morning Shift )

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.

Correct answer is 175

$\begin{aligned} v = u + a t \\ u + 50 = u + a \Rightarrow a = 50 m / s^{2} \\ 125 = u t + \frac{1}{2} a t^{2} \\ 125 = u + \frac{a}{2} \\ \Rightarrow u = 100 m / s \\ ∴ S_{n^{4 t}} = u + \frac{a}{2} [2 n - 1] \\ = 175 m \end{aligned}$

7. (JEE Main 2023 (Online) 13th April Morning Shift )

Two trains 'A' and 'B' of length ' $l$ ' and ' $4 l$ ' are travelling into a tunnel of length ' $L$ ' in parallel tracks from opposite directions with velocities $108 km / h$ and $72 km / h$ , respectively. If train 'A' takes $35 s$ less time than train 'B' to cross the tunnel then. length ' $L$ ' of tunnel is :

(Given $L = 60 l$ )

A. 900 m

B. 1200 m

C. 1800 m

D. 2700 m

Correct Option is (C)

Let's start by converting the velocities of both trains to m/s:

Train A: $108 \frac{k m}{h} \times \frac{1000 m}{k m} \times \frac{1 h}{3600 s} = 30 \frac{m}{s}$

Train B: $72 \frac{k m}{h} \times \frac{1000 m}{k m} \times \frac{1 h}{3600 s} = 20 \frac{m}{s}$

To cross the tunnel, Train A has to cover a distance equal to the length of the tunnel plus its own length: $L + l$

Similarly, Train B has to cover a distance equal to the length of the tunnel plus its own length: $L + 4 l$

We are given that Train A takes 35 seconds less time than Train B to cross the tunnel. Let's denote the time taken by Train A as $t_{A}$ and the time taken by Train B as $t_{B}$ . Then, we have:

$t_{B} = t_{A} + 35$

Using the formula distance = velocity × time, we can write the equations for both trains:

Train A: $(L + l) = 30 t_{A}$

Train B: $(L + 4 l) = 20 t_{B}$

Now, we can substitute $t_{B} = t_{A} + 35$ in the equation for Train B:

$(L + 4 l) = 20 (t_{A} + 35)$

We have two equations and two unknowns ( $L$ and $t_{A}$ ). We can solve this system of equations by eliminating one of the unknowns. Let's eliminate $t_{A}$ by expressing it from the equation for Train A:

$t_{A} = \frac{L + l}{30}$

Now, substitute this expression for $t_{A}$ in the equation for Train B:

$(L + 4 l) = 20 (\frac{L + l}{30} + 35)$

Multiplying both sides by 30 to get rid of the fraction:

$30 (L + 4 l) = 20 (L + l) + 20 \times 35 \times 30$

Expanding the equation:

$30 L + 120 l = 20 L + 20 l + 21, 000$

Simplify:

$10 L + 100 l = 21, 000$

Since we are given that $L = 60 l$ , substitute this into the equation:

$10 (60 l) + 100 l = 21, 000$

Solve for $l$ :

$700 l = 21, 000$

$l = 30$

Now, substitute the value of $l$ back into the equation for $L$ :

$L = 60 l = 60 \times 30 = 1800$

So, the length of the tunnel is:

1800 m

8. (JEE Main 2023 (Online) 6th April Evening Shift )

A particle starts with an initial velocity of $10.0 {ms}^{- 1}$ along $x$ -direction and accelerates uniformly at the rate of $2.0 {ms}^{- 2}$ . The time taken by the particle to reach the velocity of $60.0 {ms}^{- 1}$ is __________.

A. 30s

B. 6s

C. 3s

D. 25s

Correct Option is (D)

To find the time taken by the particle to reach the velocity of $60.0 {ms}^{- 1}$ , we can use the formula:

$v = u + a t$

Where: $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time taken.

Plugging in the given values:

$60.0 {ms}^{- 1} = 10.0 {ms}^{- 1} + 2.0 {ms}^{- 2} \cdot t$

Solve for $t$ :

$50.0 {ms}^{- 1} = 2.0 {ms}^{- 2} \cdot t$

$t = \frac{50.0 {ms}^{- 1}}{2.0 {ms}^{- 2}} = 25 s$

So, the time taken by the particle to reach the velocity of $60.0 {ms}^{- 1}$ is 25 seconds.

9. (JEE Main 2023 (Online) 1st February Evening Shift )

For a train engine moving with speed of $20 {ms}^{- 1}$ , the driver must apply brakes at a distance of 500 $m$ before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $\sqrt{x} {ms}^{- 1}$ . The value of $x$ is ____________.

(Assuming same retardation is produced by brakes)

Correct answer is (200)

By using $3^{rd}$ equation of motion

$\begin{aligned} v^{2} = u^{2} + 2 a s \\ (0)^{2} = u^{2} + 2 a s \\ u^{2} = - 2 a s \\ S = \frac{u^{2}}{2 a} - \frac{(20)^{2}}{2 \times a} = 500 \\ acceleration of the train, a = - \frac{400}{1000} = - 0.4 m / \sec \end{aligned}$

Now, if the brakes are applied at $S = 250 m$ i.e. half of the distance

$\begin{aligned} v^{2} = u^{2} + 2 a s \\ v^{2} = (20)^{2} + 2 (- 0.4) \times 250 \\ v^{2} = 400 - 2 \times \frac{4}{10} \times 250 \\ v^{2} = 200 \\ v = \sqrt{200} \\ Given \Rightarrow v = \sqrt{x} \\ ∴ x = 200 \end{aligned}$

10. (JEE Main 2022 (Online) 27th July Evening Shift )

The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is :

A. 2.0

B. 1.0

C. 0.5

D. 1.5

Correct Option is (C)

S = 4 cm

$v_{4}^{'} = \frac{v}{3}$ , a = constant

$v_{4 + x} = 0$

$(v^{2} - \frac{v^{2}}{a}) = 2 a (4)$

$(v^{2} - 0) = 2 a (4 + x)$

$\frac{4}{4 + x} = \frac{8}{9}$

$\Rightarrow x = 0.5$ m