Question
A student, intending to catch a bus on the bus stand, finds that the bus starts and accelerates away from him/her with uniform acceleration $a = 2{\text{ }ms}^{- 2}$ when he is 16 m far from the bus. he runs fast enough with a uniform speed so as just to catch the bus. The minimum speed with which the student must run, is
Options
Solution
Sol. Key concept: Relative motion and the limiting condition for catching an accelerating body.
Take the student’s starting point as the origin. After time t, the student has travelled vt, whereas the bus is at 16 + at²/2. Catching the bus requires these positions to be equal.
$$vt = 16 + \frac{at^{2}}{2}$$
This quadratic in t must have a real solution. The minimum running speed occurs when its discriminant is zero, meaning that the student just catches the bus.
$$v_{\min} = \sqrt{2a(16)} = \sqrt{2 \times 2 \times 16} = 8\ m\ s^{- 1}$$
At this limiting instant, t = v/a = 4 s; the student and bus both have speed 8 m s⁻¹. Hence option c.
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