1.

A train starts from rest and accelerates uniformly at 100 m "minute"^(-2) for 10 minutes. Find the velocity acquired by the train. It then maintains a constant velocity for 20 minutes. The brakes are then applied and the train is uniformly retarded. It comes to rest in 5 minutes. Draw a velocity-time graph and use it to find : (i) the retardation in the last 5 minutes, (ii) total distance travelled, and (iii) the average velocity of the train.

Answer»

Solution :INITIAL velocity = 0, time interval = 10 minute, acceleration = 100 m `"minute"^(-2)`.
Acceleration = `("Final velocity - Initial velocity ")/("Time interval ") `
`= ("Final velocity"-0)/("Time interval") `
or Final velocity = acceleration `xx` time interval
= 100 m `"minute"^(-2) xx 10 ` minute
= 1000 m `"minute"^(-1)`
`:.` The final velocity acquired = 1000 m `"minute"^(-1)`
The velocity - time graph is shown in Fig 2.28.

(i) Retardation in the last 5 minutes
= - slope of the line BC.
` = - (BE)/(EC) = - ((0-1000)"m minute"^(-1))/((35-30)"minute" )`
`= - (-1000 "m minute"^(-1))/(5 "minute") = 200m"minute"^(-2)`
(ii) Total distance travelled
= Area of trapezium OABC
`= (1)/(2) (OC + AB) xx AD`
`= (1)/(2) (35 +20) ` minute `xx1000` m `"minute"^(-1)`
`= 55 xx500` m
= 27500m ( or 27.5 km ) .
(III) Average velocity = `("Total distance travelled ") /("Total time of travel")`
`= (27500m)/(35 "minute ")=785.7 "m minute"^(-1)` .


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