1.

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 graph. Calculate a) the acceleration b) retardation c) total distance travelled by the train

Answer»

Hey dear.....

First let us plot the graph with the data given.

Let the starting point be the ORIGIN in the speed - time graph.




(a) Uniform rate of acceleration = slope of AB
= (20−0) (m/s)/ [(200−0) s]
= 0.1 m/s2.

(b) Uniform retardation = − slope of CD

= − (0−20)(m/s) /[(800−700)s]

= 0.2 m/s2.

(c) Total distance COVERED before stopping = area under the curve ABCD
= area of ΔABE + area of rectangle BCFE + area of ΔCDF
= ½ x 200 x 20 + 500 x 20 + ½ x 100 x 20 m
= 2000 + 10000 + 1000 m
= 13,000 m = 13 km

(d) Average speed = Total distance covered/total time taken = 13000 m /(800 s)
= 16.25 m/s

HOPE IT HELPS ✌ ✌ ✌



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