Question

Q.4. A train starting from rest, picks up speed of 20 ms 1 in 200s, it continues to move at same rate
for the next 500s and then it is brought to rest in another 100 s.
a) Plot a speed time graph
b) From the graph calculate:
(i) uniform rate of acceleration
(ii) total distance covered before stopping
(iii) Average speed over total journey (iv) acceleration between 200s to 700s​

Answer 1

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 /(800s)

= 16.25 m/s

Explanation: