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(\rm m)/(\rm s)\right)^2 = 2\left(3(\rm m)/(\mathrm s^2)\right)(120\,\mathrm m)`

Solve for v :

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