JEE PYQ: Motion in a Straight Line - Question ID 63a001282c8f (JEE Main 2021)

ID: 63a001282c8fJEE Main 2021Single Correct MCQ
An engine of a train, moving with uniform acceleration, passes the signal-post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is :
JEE Question illustration 63a001282c8f

Select Option

Step-by-step Explanation

Core Formula & Concept:

In problems involving uniformly accelerated motion in a straight line, the following kinematic relations are fundamental:

  • Velocity–displacement relation (no time involved): v2u2=2asv^2 - u^2 = 2\,a\,s where
    uu = initial velocity,
    vv = final velocity,
    aa = uniform acceleration,
    ss = displacement.
  • Velocity at any intermediate point can be found by applying the same relation to the segment of interest.

Here the train’s engine (front) passes the signal post with speed uu and the last compartment (rear) passes it with speed vv. We are to find the speed of the middle point of the train as it passes the same signal post.

Step-by-Step Derivation:

1. Let the total length of the train be LL. 2. When the engine (front) passes the signal post, its velocity is uu. 3. When the last compartment (rear) passes the same post, the engine has moved forward by distance LL and its velocity has increased to vv. 4. Apply the velocity–displacement relation over the entire train length LL: v2u2=2aLaL=v2u22.v^2 - u^2 = 2\,a\,L \quad\Longrightarrow\quad a\,L = \frac{v^2 - u^2}{2}. 5. Now consider the middle point of the train. As the engine moves from the signal post to the position where the middle point reaches the post, the engine covers a distance L2\tfrac{L}{2}. 6. Let ww be the velocity of the middle point when it passes the post. Applying the same kinematic relation over the half-length L2\tfrac{L}{2} gives w2u2=2a(L2)=aL.w^2 - u^2 = 2\,a\,\Bigl(\tfrac{L}{2}\Bigr) = a\,L. 7. Substitute aLa\,L from step 4: w2u2=v2u22w2=u2+v2u22=v2+u22.w^2 - u^2 = \frac{v^2 - u^2}{2} \quad\Longrightarrow\quad w^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{v^2 + u^2}{2}. 8. Therefore the required velocity is w=v2+u22.w = \sqrt{\frac{v^2 + u^2}{2}}.

Common Traps & Exam Tip:

• Many students mistakenly take the arithmetic mean u+v2\tfrac{u+v}{2} (option A), forgetting that velocity under uniform acceleration does not vary linearly with distance.
• Others confuse the velocity–displacement relation with the velocity–time relation and arrive at incorrect expressions.
Exam Tip: Always write down the kinematic relation that connects the known quantities (here v2u2=2asv^2 - u^2 = 2\,a\,s) and then apply it to the segment of interest.

Related Questions from Motion in a Straight Line

ID: 3bcc2581db85JEE Main 2026

A gas balloon is going up with a constant velocity of 10 m/s10 \mathrm{~m} / \mathrm{s}. When this balloon reached a height of 75 m , a stone is dropped from it and balloon keeps moving up with the same velocity. The height of the balloon when the stone hits the ground is ____\_\_\_\_ m. (Take g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^2 )

View Solution →
ID: 2e82840fda73JEE Main 2026

The velocity (v)(v) versus time (t)(t) plot of a particle is shown in the figure, for a time interval of 40 s . The total distance travelled by the particle and the average velocity during this period are, respectively

____\_\_\_\_.

JEE Main 2026 (Online) 5th April Evening Shift Physics - Motion in a Straight Line Question 3 English
View Solution →
ID: f97d835af5d4JEE Main 2026

Two cars AA and BB are moving in the same direction along a straight line with speeds 100 km/h100 \mathrm{~km} / \mathrm{h} and 80 km/h80 \mathrm{~km} / \mathrm{h}, respectively such that car AA is moving ahead of car BB. A person in car BB throws a stone with a speed vv so that it hits the car AA with a speed of 5 m/s5 \mathrm{~m} / \mathrm{s}. The value of vv is ____\_\_\_\_ km/h\mathrm{km} / \mathrm{h}.

View Solution →
ID: 2b8b065cdd64JEE Main 2026

A particle starts moving from time t=0t=0 and its coordinate is given as x(t)=4t33tx(t) = 4t^3 - 3t

A. The particle returns to its original position (origin) 0.866 units later

B. The particle is 1 unit away from origin at its turning point

C. Acceleration of the particle is non-negative

D. The particle is 0.5 units away from origin at its turning point

E. Particle never turns back as acceleration is non-negative

Choose the correct answer from the options given below :

View Solution →