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. (a) 112.5 m and 22.5 s (b) 337.5 m and 25 s (c) 112.5 m and 15 s (d) 225.5 m and 10 s

Answer: (a). In 15 s: car = 1/2× 15× 45 = 337.5 m, scooter = 30× 15 = 450 m Difference = 450 - 337.5 = 112.5 m After 15 s the car (45 m/s) gains 15 m/s on the scooter: extra time = (112.5)/(15) = 7.5 s Catch-up at t = 15 + 7.5 = 22.5 s. So the answer is (a).

The velocity-time graphs of a car and a scooter are shown in the… | JEE Main 2018 (15 Apr, Shift 1)

Q31JEE Main 2018 · 15 APR · SHIFT 1Medium

Graphs of Motion

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 $15\text{ s}$ and (ii) the time at which the car will catch up with the scooter are, respectively.

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Answer: (a)

In 15 s: car $= \frac{1}{2}\times 15\times 45 = 337.5\text{ m}$ , scooter $= 30\times 15 = 450\text{ m}$
Difference $= 450 - 337.5 = 112.5\text{ m}$
After 15 s the car (45 m/s) gains $15\text{ m/s}$ on the scooter: extra time $= \frac{112.5}{15} = 7.5\text{ s}$
Catch-up at $t = 15 + 7.5 = 22.5\text{ s}$ . So the answer is (a).

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