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

Using equation, \(a=\frac{v-u}{t}\) and
\(s=u t+\frac{1}{2} a t^2\)
Distance travelled by car in 15 sec
\(=\frac{1}{2} \frac{(45)}{15}(15)^2\)
\(=\frac{675}{2} m\)
Distance traveled by scooter in 15 seconds
\(=30 \times 15=450 (\text {distance}=\text {speed} \text { × } \text {time})\)
Difference between distance travelled by car and scooter in 15 sec ,
\(450-337.5=112.5 m\)
Let car catches scooter in time \(t\);
\(\frac{675}{2}+45(t-15)=30 t\)
\(337.5+45 t-675=30 t\)
\(\Rightarrow 15 t=337.5\)
\(\Rightarrow t=22.5 sec\)