Question

A train starting from rest picks up speed of 10m/s in 100 sec. It continues to move at the same speed for the next 250 sec. It is then brought to rest in the next 50 sec. Plot a speed-time graph for the entire motion of the train. Calculate:
A) acceleration of the train while accelerating
B) retardation of the train while retarding
C) the total distance covered by the train.

Answer 1

A=v-u/t
A=10/100
= 1/10m/sec²

-a = u-v/t
= 10/50
=1/5 m/sec²

Answer 2

Answer:

The acceleration of the train will be $0.1 m/s^2$, the retardation of the train  will be $-0.2\ m/s^2$ and the total distance covered by the train is 3250 m.

Explanation:

Given that,

Speed $v_{1}=10\ m/s$

Time $t_{1}=100\ sec$

Time $t_{2}=250\ sec$

Time $t_{3}=50\ sec$

(A). The acceleration of the train will be

$OA=a= \dfrac{v_{2}-v_{1}}{t}$

$a=\dfrac{10-0}{100-0}$

$a=0.1\ m/s^2$

(B). The retardation of the train  will be

$BC=a=\dfrac{10-0}{350-400}$

$a= -0.2\ m/s^2$

(C). The total distance covered by the train is

The distance = area of OABC  

$s= \dfrac{1}{2}\times(250+400)\times10$

$s= 3250\ m$

Hence, The acceleration of the train will be $0.1 m/s^2$, the retardation of the train  will be $-0.2\ m/s^2$ and the total distance covered by the train is 3250 m.