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
2,


distance s = 50 m


From the equation of motion,


s=ut + at2


50=0+(1/2)9t2


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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