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.
Answer: (A) $112.5\ \text{m}$ and $22.5\ \text{s}$
Both start from the same point at $t = 0$. The car accelerates uniformly from rest to $45\ \text{m s}^{-1}$ in $15\ \text{s}$ and then moves uniformly; the scooter moves at a constant $30\ \text{m s}^{-1}$.
**(i)** Distances in the first 15 s are the areas under the graphs:
$$s_\text{car} = \tfrac12(15)(45) = 337.5\ \text{m}, \qquad s_\text{scooter} = 30 \times 15 = 450\ \text{m}$$
Difference $= 450 - 337.5 = 112.5\ \text{m}$ (the scooter is ahead).
**(ii)** After $t = 15\ \text{s}$ the car gains on the scooter at $45 - 30 = 15\ \text{m s}^{-1}$, so it needs
$$\frac{112.5}{15} = 7.5\ \text{s}$$
more. The car catches up at $t = 15 + 7.5 = 22.5\ \text{s}$.
Solution by Sreeraj P, M.Sc Physics