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A goods train accelerating uniformly on a straight railway track, approaches an electric pole standing on the side of track. Its engine passes the pole with velocity $$u$$ and the guard's room passes with velocity $$v$$. The middle wagon of the train passes the pole with a velocity.
Let the total length of the train be $$L$$. Therefore, the middle wagon sits exactly at a distance of $$\frac{L}{2}$$ from both the engine (front) and the guard's room (back). The train is moving with a constant uniform acceleration $$a$$.
Using the third equation of motion ($$v^2 = u^2 + 2as$$) for the entire length of the train as it passes the pole:
$$v^2 = u^2 + 2aL \implies 2aL = v^2 - u^2 \quad \text{--- (Eq. 1)}$$
Let $$v_m$$ be the instantaneous velocity of the train at the exact moment its middle wagon passes the pole. Over this interval, the train has covered a displacement of exactly $$s = \frac{L}{2}$$ past the pole.
Applying the third equation of motion over this half-length interval:
$$v_m^2 = u^2 + 2a\left(\frac{L}{2}\right)$$
$$v_m^2 = u^2 + aL \quad \text{--- (Eq. 2)}$$
From Equation 1, we can isolate the term $$aL$$:
$$aL = \frac{v^2 - u^2}{2}$$
Substitute this expression for $$aL$$ directly back into Equation 2:
$$v_m^2 = u^2 + \frac{v^2 - u^2}{2}$$
$$v_m^2 = \frac{2u^2 + v^2 - u^2}{2}$$
$$v_m^2 = \frac{u^2 + v^2}{2}$$
Taking the positive square root to find the actual velocity value:
$$v_m = \sqrt{\frac{u^2 + v^2}{2}}$$
Concept Check: Because kinetic energy scales with the square of velocity ($$v^2$$), the velocity at the spatial midpoint under constant acceleration matches the root-mean-square (RMS) value of the initial and final speeds rather than a simple arithmetic average.
Correct Option Key: Option D ($$\sqrt{\frac{u^2 + v^2}{2}}$$)
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