The velocity-time graph line motion is shown in Fing. 2 (b).7. Find (a) the distance travelled and (b) displacement, between `5 and 40 s`. (c ) Acceleration betwwn `15 to 25 seconds.

Explanation

`(a) Distance travelled by body in time `5 ` to `15 ceconds =Area A_(1) ABB, _(1) =((2+4) xx 100/2 =30 m`
Distance travelled by body in time `15 to 25`
`seconds =Area BB_(1)C =(5xx2)/2 =5 m`
Distance travelled by body in time `30 to 40`
`Seconds =Area D_(1)DEF =((2+4 xx 10)/2 =30 m`
:. Total distance travelle between `5 to 40 seconds =30 + 20 +5+ 30 =85 m`
(b) Here, the ares `CDD_(1) and area `D_(1)DEF` being below the time axis will show negative dissplacement.
Here, Tota displacement between `5 to 40 seconse `=30 +20 -5- 30 15 m`
Acceleeration of the the body in time interval `15 s to 25 s`is the slope of the line `BC`
`=(-BB_(1))/(CB_(1)) =(-4)/((25 -15)) ]=(-4)/(10) =- 0.4 m//s^(2)`.