Question

A train starts from rest and moves with a constant acceleration of 2.0 Meyer per second square for half a minute. The brakes are then applied and the train comes in one minute. Find a) the total distance moved by the train,b) the maximum speed attained by the train c) the position (s) of the train at half the maximum speed​

Answer 1

Answer:

Explanation:

a) u = 0, a = 2 m/s^2, t = 30 s

Total distance = ut + 1/2 at^2

= 1/2 at^2

= 1/2 * 2 * 30 * 30

= 30 * 30 = 900 m

b) Maximum speed attained is given by the formula

v^2 - u^2 = 2as

u = 0

= v^2 = 2as

v^2 = 2 * 2 * 900

v^2 = 3600

v = $\sqrt{3600}$

v = 60 m/s