Question

a train starting from rest picks up a speed of 20 metre per second in 200 sec. It continuous to move at the same rate of 500 sec and is then brought to rest in another 100 sec plot a Speed time graph
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Answer 1

Answers

A=v-u/t

A=10/100

= 1/10m/sec²

-a = u-v/t

= 10/50

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.

Hope it helped u

Answer 2

Answer:

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