A train starting from rest, picks up a velocity of 20m/s in 200s. It continues to move at the same rate for the next 500s and is then brought to rest in another 100s. Plot a v-t graph and calculate average velocity.
Answers
Answer:
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
Answer:
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
