As soon as a car starts from rest in a certain dirction, a scooter moving with a uniform speed overtakes the car. Their velcity-time graphs are shown in Fig. 2 (b) .19.` Calcutate
(a) the difference between the distances travelled by the car and the scooter in `25 s`.
(b) the time when car will catch up the secooter.
(c ) the distance of the car and scooter from the starting point at the meeting point.
.

Explanation

(a) Dstance travelled by the car in
`=Area (OAF) + Area (ABGF) `
`= 1/2 xx 15 xx 60 xx (25 -15)`
`= 45 0 + 600 =1050 m`
Distance travelled by the scooter in `25 s`
`= 25 xx 40=1000 m`
Differnce in the distances travelled
`=1050 -1000 =50 m`
(b) The distance travelled by car in `15 s`
` =are OAF =1/2 xx 15 xx 60 =450 m`
Distance trav elled by scooter in `15 s`
`=40 xx 15 =600 m`
Let the car catch up the scooter intime (15 +t) s
After 15 second, the velcity of car isconsstant. The distance travelled by car intime (15+t) s is `
=(4 50 +50 t) m`
Distance travelled by scooter in time (15 + t)` is
`=40 (15 +t) m`
The car will catch up thecsooter, when
(450 + 60 t) =40 (15 +t) =600 + 40 t`
or ` t=7.5 s`
:. Total time, the car will catch up the scooter
`=15 + 7.5 =22 .5 s`
(C ) Distance of car at meeting point distance of scooter fromstarting point in tme `22.5 s`
`=22.5 xx 40 =900 m`.