Question

15. A train starting from rest picks up a speed of 10 m st
in 100 s. It continues to move at the same speed for
the next 250 s. It is then brought to rest in the next 50 s.
Plot a speed-time graph for the entire motion of the
train
(1) acceleration of the train while accelerating
(ii) retardation of the train while retarding
(iii) and the total distance covered by the train.
(SEPT 2011)​

Answer 1

Answer:

acceleration in first 100 seconds

= (10-0)/100= 0.1 m sec^-2

distance traveled during first 100 seconds

is S= 0.x10+ 1/2×0.1x10^4= 500 m = 0.5 Km. (a)

and Velocity it reaches in 100 secs is

V = 0+ 0.1x 100= 10 m/sec= 36 Km/ Hr.

then it covers : 250×10= 2500 m or 2.5 Km. (b)

decelerating in 50 second : a= 0-10/50= -0.2 ( negative becuase of speed reduction and NOT that train will move back !)

during these 50 seconds train will cover

Sd= 10x50- 1/2×0.2×2500= 250 m = 0.25 KM. (c)

Total distance = a +b+c= 3.25 KM.