Question

As soon as a car just starts from rest in a certain direction, a scooter moving with a uniform speed overtakes the car. Their velocity-time graph is shown. Calculate
a. The difference between the distances traveled by car and the scooter in $15s$.
b. The distance of car and scooter from the starting point at that instant.

Answer 1

a. The distance travelled by car in $15s$
$=$ Area of AOAC
$=\frac{1}{2}$ $\times$15$\times 45=337.5m$
Distance travelled by scooter in 15 s
$=$ Area of rectangle OCEF
$=15\times$ $30=450m$
Thus, the difference between distance traveled by them $=450m-337.5m=112.5m$

b. Let after time $t$ from start car will catch, up the scooter. In time $t$, the distance traveled by them is equal.
Distance travelled by car $=\frac{1}{2}$ $\times 15$ $\times 45$ $+45(t-15)$
Distance travelled by scooter $=30t$
$\frac{1}{2}$ $\times$ $15$ $\times$ $45+45(t-15)=30t$
which gives $t=22.5s$
Distance travelled by car or scooter in $22.5s=30\times 22.5$
$=675m$
So the car catches the scooter when both are at $675m$ from the starting point.