Question

A particle moves on straight line according to the velocity-time
graph shown in fig. Calculate -
1) Total distance covered
2) Average speed
(3) In which part of the graph the acceleration is maximum and also find its value.
(iv) Retardation​

Answer 1

1)In Velocity-time graph area of figure= displacement

Which is ....2+4+2+4+25= 37 m(distance)

Note: To calculate displacement just subtract The negative velocity from the positive one....

2) AVG speed = Total distance/total time

= 37/10

= 3.7m/s

3)Acceleration= v-u/t

To find highest acc..

Therefore note the biggest velo gap in unit time.......

Which is from 4 to 5 second

Now 10-2/1

=8m/sec^2

4)Retardation is present in the last 5 seconds of the graph...

To calculate it use v-u/t

0-10/5

=-2m/sec^2

I HOPE U UNDERSTAND!!:)