As soon as a car just starts from rest in a certain dercation, a scooter moveing with a uniform speed overtakes the car. Their velocity-time graph is shown in . Calculate
.
a. The difference between the distances travlled by the car and the scooter in `15 s`,
b. Find the time at which the car passes the scooter and the distance of car and scooter from the starting point at that instant.

Options

A. `110 m` `22.5 s` `600 m`
B. `112.5 m` `25 s` `675 m`
C. `112.5 m` `22.5 s` `675 m`
D. `1125 m` `22.5 s` `675 m`

Answer

C

Explanation

a. The distance travekked by car in `15 `s
`=Area` of `Delta OAC`
`=(1)/(2) xx 15 xx 45 =337. m`
Distance travelled by scooter in `15 s`
`=Area` of retangle `OCEF`
`=15xx30=450 m`
Thus, difference between distance travelled by them
`=450 m-337.5 m= 112.5 m`
b. Let aftre time `t` from start car will catch up the scooter. In time `t`, the distance travelled by them are equal.
Distance travelled by car `=(1)/(2)xx 1545+ 45(t-15)`
Distance travelled by scooter `=30t`
`(1)/(2) xx 15 xx 45 +45 (t-15)=30t`
which gives `t=22.5 s`
Distance travelled by car or scooter in `22.5 s=30xx22.5`
`=675 m`
So the catr catches the scoter when both are at `67.5 m` from the starting point.