question_answer

A passenger is standing 'd' metres away from a bus. The bus begins to move with constant acceleration a. To catch the bus, the passenger runs at a constant speed \[\upsilon \]towards the bus. What must be the minimum speed of the passenger so that he may catch the bus.

Answer:

                Let the passenger catches the bus after time t. Distance travelled by the bus in time t,                 \[{{s}_{1}}=0+\frac{1}{2}a{{t}^{2}}=\frac{1}{2}a{{t}^{2}}\]                                                                                                                             ... (i) Distance travelled by the passenger, \[{{s}_{2}}=\upsilon t+0=\upsilon t\]                          ...(ii) The passenger will catch the bus if,          \[d={{s}_{1}}={{s}_{2}}\]                               or \[d+\frac{1}{2}a{{t}^{2}}=\upsilon t\]                or            \[\frac{1}{2}a{{t}^{2}}-\upsilon t+d=0\] or \[t=\frac{\left[ \upsilon \pm \sqrt{{{\upsilon }^{2}}-2ad} \right]}{a}\]. The passenger will catch the bus if t is real , i.e., \[{{\upsilon }^{2}}\ge 2ad\] or \[\upsilon \ge \sqrt{2ad}\] Thus, minimum speed of passenger for catching the bus is \[\sqrt{2ad}\].