At the moment t=0 particle leaves the origin and moves in the positive direction of the x-axis. Its velocity varies with time as `v=10(1-t//5)`. The dislpacement and distance in 8 second will be

Options

A. 16 m, 34 m
B. 16 m, 25 m
C. 16 m, 16 m
D. 16 m, 9m

Answer

A

Explanation

`v=10(1-t/5)=10-2t`
`(ds)/dt =10-2t`
`int_(0)^(s) ds=int_(0)^(8)(10-2t)dt`
`s=10t-2t^(2)/2=|10t=t^(2)|_(0)^(8)`
`=10(8)-(8)^(2)=80-64=16 m`
Displacement`=16 m`
To find the distance, first determine whether velocity becomes zero in the given time interval or not.
`v=10-2t=0impliest=5 s`

Distance`=int_(0)^(5) v dt+|int_(5)^(8) v dt|`
`=25+|-9|=25+9=34 m`
OR
Draw a v-t graph

Area of v-t graph gives displacement and distance
Displacement`=25-9=16 m`
Distance`=25+9=34 m`