Question

A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of $18{\mathrm{ms}}^{-1}$when it passes a railway worker who is standing 180 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See below figure)

Short answer

The velocity of the last car when it passes the worker will be $21.42{\mathrm{ms}}^{-1}$.

Step-by-step solution

Step 1. Meaning of uniform motion

Uniform acceleration denotes a constant rate of velocity change with time.

Step 2. Data identification and assumption

Length of the train,$L=75\mathrm{m}$

Distance traveled when the front car reaches the man,$d=180\mathrm{m}$

Speed of the train when it reaches the man,$v_1=18{\mathrm{ms}}^{-1}$

The initial velocity of the train,$u=0{\mathrm{ms}}^{-1}$

Let the acceleration of the train be a.

Step 3. Substitution of values in the third equation of motion

When the front of the train reaches the worker, the displacement of the train is d, and the velocity is $v_1$.

Substituting these values in the third equation of motion, you get.

$\begin{array}{c}{v}_1^2-u^2=2ad\\ {18}^2-0^2=2a\times 180\end{array}$

From the above equation,

$a=0.9{\mathrm{ms}}^{-2}$

Step 4. Calculation of the speed when the last car passes

Let the velocity when the last car passes be $v_2$. At this instance, the displacement of the train from its initial position is d + L.

Substituting this value in the third equation of motion,

$v_2^2-u^2=2a\left(d+L\right)$

Substituting the values of a, u, L, and d in the above equation,

$v_2^2-0^2=2\times 0.9\times \left(180+75\right)$

Solving the above equation,

role="math" localid="1642844457086" $v_2=21.42{\mathrm{ms}}^{-1}$

Thus, the velocity of the last car when it passes the man will be $21.42{\mathrm{ms}}^{-1}$.