The velocity-time graphs of a car and a scooter starting from the same point are shown in the figure. (i) the difference between the distance travelled by the car and the scooter in 15s and (ii) the time at which the car will catch up with the scooter are, respectively

Using equation,$a=\frac{v-u}{t}$  and  $S=ut+\frac{1}{2}a{t}^2$
Distance travelled by car in 15 sec   $=\frac{1}{2}\frac{\left(45\right)}{15}{\left(15\right)}^2=\frac{675}{2}m$
Distance travelled by scooter in 15 seconds  $=30\times 15=450$
$\left(\therefore \text{distance}=\text{speed}\times \text{time}\right)$
Difference between distance travelled by car and scooter in 15 sec, 450-337.5 = 112.5m
Let car catches scooter in time t; $\frac{675}{2}+45\left(t-15\right)=30t\Rightarrow 337.5+45t-675=30t\Rightarrow 15t=337.5$

$\Rightarrow t=22.5\sec$