Discussion
Two cars P and Q are travelling on a straight path and are 60 m apart as shown in the figure; Car P is moving with a constant velocity of 36 kmph, while car Q is moving at a constant velocity of 18 kmph. At this instant, the driver in car P applies the brake and collision occurs with car Q after 30 seconds.
[The figure shows car P behind car Q, both moving to the right, P at 36 kmph and Q at 18 kmph, with a 60 m gap between them.]
Assuming uniform deceleration due to braking, which one of the following is the CORRECT velocity (in m/s) of the car P just before the collision?
Answer: D
Car P hits car Q at 4 m/s.
Convert the speeds: 36 kmph = 10 m/s and 18 kmph = 5 m/s. In 30 s car Q moves 5 × 30 = 150 m, so car P must cover the 60 m gap plus 150 m, that is 210 m, in the same 30 s with uniform deceleration a:
\[ \begin{aligned} 210 &= 10(30) - \tfrac{1}{2}a(30)^{2} \\ 450a &= 90 \Rightarrow a = 0.2\ \text{m/s}^{2} \\ v &= 10 - 0.2 \times 30 = 4\ \text{m/s} \end{aligned} \]
Check: the average speed of P is (10 + 4)/2 = 7 m/s, and 7 × 30 = 210 m, as required.
5 m/s is car Q's speed, which P would have only if it just touched Q without closing on it. 16 m/s is faster than P's starting speed, impossible under braking, and 1 m/s does not satisfy the distance condition.
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