Question

A train starting from rest , pick up a speed of 10m/s in 100s . It continues to move at the same speed for the next 250s . It is then brought to rest in the next 50s . Plot a speed - time graph for the entire motion of the train. (i) acceleration of the train while accelerating (ii) retardation of the train while regarding (iii) and the total distance covered by the train.

Answer 1

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Answer 2

Answer: The acceleration is $0.1 m/s^2$, the retardation is $-0.2 m/s^2$ and the total distance is 3250 m.

Explanation:

Given that,

Speed v = 10 m/s

Time t = 100 s

According to graph,

(I). The acceleration of the train is

Acceleration =change in velocity/time

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

$a = 0.1 m/s^2$

(II). The retardation of the train is

Retardation=change in velocity/time

$a = \dfrac{0-10}{50}$

$a = -0.2 m/s^2$

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

According to speed- time graph

Area under the curve gives total distance

Total distance = Area of triangle OAB + area ABCD +area of triangle DCE

$Distance= \dfrac{1}{2}\times100\times10+250\times10+\dfrac{1}{2}\times50\times10$

$Distance = 3250 m$

Hence, The acceleration is $0.1 m/s^2$, the retardation is $-0.2 m/s^2$ and the total distance is 3250 m.