JEE Main Motion in Straight Line PYQ

1. (JEE Main 2024 (Online) 1st February Evening Shift )

Train A is moving along two parallel rail tracks towards north with speed $72 km / h$ and train B is moving towards south with speed $108 km / h$ . Velocity of train B with respect to A and velocity of ground with respect to B are (in ${ms}^{- 1}$ ):

A.-50 and -30

B.-50 and 30

C.-30 and 50

D.50 and -30

Correct option is (b)

To find the velocity of Train B with respect to Train A, we have to subtract the velocity of Train A from the velocity of Train B, keeping in mind that they are moving in opposite directions. Since they are moving in opposite directions, the relative velocity is calculated by adding their magnitudes when converting into the same unit, which in this case is meters per second.

First, let's convert the speeds from km/h to m/s by multiplying by the conversion factor $\frac{1000 m/km}{3600 s/h} = \frac{5}{18} m/s$ .

For Train A:

v_{A} = 72 \frac{km}{h} \times \frac{5}{18} \frac{m/s}{(km/h)} = 20 \frac{m}{s}

For Train B:

v_{B} = 108 \frac{km}{h} \times \frac{5}{18} \frac{m/s}{(km/h)} = 30 \frac{m}{s}

To find the velocity of B relative to A ( $v_{B / A}$ ), we consider the direction: Train B is moving towards the south and Train A is moving towards the north. Therefore, relative to Train A, Train B is moving even faster towards the south, we calculate:

v_{B / A} = v_{B} + v_{A} = 30 \frac{m}{s} + 20 \frac{m}{s} = 50 \frac{m}{s}

Since Train B is moving towards the south and Train A towards the north, we take the southward direction as negative in our coordinate system for this calculation. That means the velocity of B with respect to A is:

v_{B / A} = - 50 \frac{m}{s}

Next, we calculate the velocity of the ground with respect to Train B ( $v_{g r o u n d / B}$ ). The ground is stationary, thus it has a velocity of 0 m/s in any direction. The velocity of an object with respect to another object moving is just the opposite of the second object's velocity. Thus:

v_{g r o u n d / B} = - v_{B} = - 30 \frac{m}{s}

However, because we are considering the southward direction as negative, the negative of a southward velocity is a northward velocity. Hence we get:

v_{g r o u n d / B} = 30 \frac{m}{s}

So the velocity of Train B with respect to Train A is -50 m/s, and the velocity of the ground with respect to Train B is 30 m/s. Therefore, the correct answer is:

Option A: -50 m/s and -30 m/s. (Incorrect, because the velocity of ground with respect to B is positive in our chosen coordinate system)

Option B: -50 m/s and 30 m/s. (Correct)

Option C: -30 m/s and 50 m/s. (Incorrect)

Option D: 50 m/s and -30 m/s. (Incorrect)

Thus, the correct answer is Option B: -50 m/s and 30 m/s.

2 (JEE Main 2023 (Online) 13th April Evening Shift )

A passenger sitting in a train A moving at $90 km / h$ observes another train $B$ moving in the opposite direction for $8 s$ . If the velocity of the train B is $54 km / h$ , then length of train B is:

A. 80 m

B. 200 m

C. 120 m

D. 320 m

Correct Option is (D)

To find the length of train B, we first need to determine the relative velocity between train A and train B. Since they are moving in opposite directions, their velocities add up:

$v_{A B} = v_{A} + v_{B} = 90 km / h + 54 km / h = 144 km / h$

Now, we need to convert this relative velocity to meters per second:

$v_{A B} = \frac{144 km / h \times 1000 m / km}{3600 s / h} = 40 m / s$

The passenger in train A observes train B for 8 seconds. To find the length of train B, we can use the formula:

$length = relative velocity \times time$

$length = 40 m / s \times 8 s = 320 m$

So, the length of train B is 320 meters.

3 (JEE Main 2021 (Online) 20th July Evening Shift )

A boy reaches the airport and finds that the escalator is not working. He walks up the stationary escalator in time t₁. If he remains stationary on a moving escalator then the escalator takes him up in time t₂. The time taken by him to walk up on the moving escalator will be :

A. $\frac{t_{1} t_{2}}{t_{2} - t_{1}}$

B. $\frac{t_{1} + t_{2}}{2}$

C. $\frac{t_{1} t_{2}}{t_{2} + t_{1}}$

D. $t_{2} - t_{1}$

Correct Option is (C)

L = Length of escalator

$V_{b / e s c} = \frac{L}{t_{1}}$

When only escalator is moving.

$V_{e s c} = \frac{L}{t_{2}}$

when both are moving

$V_{b / g} = V_{b / e s c} + V_{e s c}$

$V_{b / g} = \frac{L}{t_{1}} + \frac{L}{t_{2}} \Rightarrow [t = \frac{L}{V_{b / g}} = \frac{t_{1} t_{2}}{t_{1} + t_{2}}]$

4 (JEE Main 2020 (Online) 2nd September Morning Slot )

Train A and train B are running on parallel tracks in the opposite directions with speeds of 36 km/hour and 72 km/hour, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/ hour. Speed (in ms^–1) of this person as observed from train B will be close to :
(take the distance between the tracks as negligible)

A. 30.5 ms^–1

B. 29.5 ms^–1

C. 31.5 ms^–1

D. 28.5 ms^–1

Correct Option is (B)

Velocity of man with respect to ground

${\vec{V}}_{m / g}$ = ${\vec{V}}_{m / A}$ + ${\vec{V}}_{A}$

= -1.8 + 36

Velocity of man w.r.t. B

${\vec{V}}_{m / B}$ = ${\vec{V}}_{m}$ - ${\vec{V}}_{B}$

= –1.8 + 36 – (–72)

= 106.2 km/hr

= 29.5 m/s

5 (JEE Main 2019 (Online) 12th January Morning Slot )

A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when : (i) they are moving in the same direction , and (ii) in the opposite direction is :

A. $\frac{25}{11}$

B. $\frac{3}{2}$

C. $\frac{11}{5}$

D. $\frac{5}{2}$

Correct Option is (C)

The total distance to be travelled by the train is 60 + 120 = 180 m.

When the trains are moving in the same direction, relative velocity is
v₁ – v₂ = 80 – 30 = 50 km hr^–1

So time taken to cross each other,

t₁ = $\frac{180}{50 \times \frac{10^{3}}{3600}}$

When the trains are moving in opposite direction, relative velocity is
|v₁ – (–v₂ )| = 80 + 30 = 110 km hr^–1

$∴$ Time taken to cross each other

t₂ = $\frac{180}{110 \times \frac{10^{3}}{3600}}$

$∴$ $\frac{t_{1}}{t_{2}} = \frac{\frac{180}{50 \times \frac{10^{3}}{3600}}}{\frac{180}{110 \times \frac{10^{3}}{3600}}}$ = $\frac{11}{5}$