A particle starts from rest with a fconstant acceleration. At a time t second, the speed is found to be 100m/s and one second later the speed becomes 150m/s. find (a) the acceleration and (b) the distance travelled during the (t+1)th second.

(a) Velocity at time t is 100m/s =a.(t second).............................1

and velocity at time (t+1) second is  150m/s=a. (t+1) second....................2

(b) consider the interval t second to (t+1) second, time elapsed =1s

Initial Velocity = 100 m/s

Final velocity = 150 m/s.

Thus,  (150 m/s)² = (100m/s)² + 2(50 m/s² )x

or,                        x= 125 m.

This is a simply question that can be solved in more than one method. Given, Let velocity at the instant time 't' = 100 m/s now velocity after 1 second that is at 't+1' sec velocity =150 Let acceleration be a. now from equations of motion, v = u +at 150 = 100 + at' ...............1 where t' = [t+1]-t =1 sec. there fore a = 50 m/s2. [a bit] distance travelled : v2 -u2 = 2as there fore,2500 m