Question

A train travel on a straight track passing signal A at 20ms-1. It accelerates uniformly a 3ms-2 and reaches signal B 120m furthe than A. At B, the velocity of the train is? ​

Answer 1

Since acceleration is uniform, if the velocity at point B is v, then

$v^2 - \left(20\dfrac{\rm m}{\rm s}\right)^2 = 2\left(3\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

Solve for v :

$v^2 = \left(20\dfrac{\rm m}{\rm s}\right)^2 + 2\left(3\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m) \\\\ v = \sqrt{\left(20\dfrac{\rm m}{\rm s}\right)^2 + 2\left(3\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)} \\\\ \boxed{v \approx 34 \dfrac{\rm m}{\rm s}}$