Question

A train accelerates from the rest at the station A at a rate of 0.8 m/s2 for 25 s, the travels at constant velocity for 77.5 s before it comes to rest at the station B after period of retardation lasting 20 s. What is the distance between stations A and B, and the average velocity of the train between the stations?

I have calculated the distance of first part via equation s=vit+1/2at^2 and got the answer 250 then i found the velocity
20 m/s and now got stuck with retardation part . Please tell me ! soon

Answer 1

after 25 seconds, v=25*.8 = 20

when it starts to decelerate, a = -1 m/s^2, since it takes 20s to decrease v to 0.

so, the total distance is

s = 1/2 .8*20^2 + 20*77.5 + 20*10 - .5*400 = 1710 m

avg speed is thus 1710/(20+77.5+20) m/s