Question

A cyclist starts from place A and covers a distance of 2400 to reach the place B. If the maximum possible acceleration of the cyclist is 2m/s^2 and its maximum retardation is 4m/s^2, what is the least time the cyclist can undertake the journey from A to B?

Answer 1

Given: A cyclist starts from place A and covers a distance of 2400 to reach the place B. The maximum possible acceleration of the cyclist is 2m/s² and its maximum retardation is 4m/s².

To find: The least time the cyclist can undertake the journey from A to B.

Solution:

• Motion in one dimension provides a formula written as follows,

        $S = ut + \frac{1}{2}at^{2}$

• Here, S is the distance travelled, u is the initial velocity, t is the time taken and a is the acceleration produced.
• Since the cyclist starts from place A, the initial velocity is zero. So, the equation is now written as,

        $S = \frac{1}{2} at^{2}$

• For the maximum acceleration, the equation can be written as,

        $S_{1} = \frac{1}{2} (2ms^{-2}) t^{2}$

• For the maximum retardation, the equation can be written as,

        $S_{2} = \frac{1}{2} (-4ms^{-2}) t^{2}$

• But the total distance covered is 2400 m, so,

        $S = S_{1} + S_{2}$

            $= 2400m$

• On rearranging and solving further, the value of t is found to be,

        $S = \frac{1}{2} t^{2} (2 + (-4) )$

        $2400 m = \frac{1}{2} t^{2} (-2)$

        $t = 20\sqrt{6} s$

Therefore, the least time the cyclist can undertake the journey from A to B is 20√6 s.