Answer :

The superman has to move with maximum possible acceleration, to reach the given distance in minimum time,

Let us consider the maximum acceleration is ‘a‘.
ma − μR = 0

ma = μ mg
a = μg = 0.9 × 10 = 9 m/s2

(a) First figure shows the first case

As per the question, the initial velocity,
initial velocity u = 0, t = ?
acceleration a = 9 m/s

distance s = 50 m

From the equation of motion,

s=ut + at2


t=10/3 s

(b) Second figure represents the second case

After covering 50 m, the velocity of the athlete is

v = u + at
=0+9×103 m/s

=30 m/s

He has to stop in minimum time. Hence, the deceleration,

a =−9 m/s2 (max)
R = mg
ma = μR
(maximum frictional force)
ma = μmg
a = μg
= 9 m/s
2 (deceleration)
u1 = 30 m/s, v = 0

t=10/3 sec

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