91Ó°ÊÓ

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

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.

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.

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

From the above equation,

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

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,

$v_2^2-0^2=2×0.9×\left(180+75\right)$

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 $21.42{\mathrm{ms}}^{-1}$.