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* − μ*R* = 0

*⇒* *ma* = μ *mg**⇒* * a = μg* = 0.9 × 10 = 9 m/s

^{2}

(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 + at^{2}

50=0+(1/2)9t^{2}

⇒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/s^{2} (max)*R* *= mgma = μR* (maximum frictional force)

*ma*= μ*mg**⇒*

*a*= μ*g*= 9 m/s

^{2}(deceleration)

*u*

_{1}= 30 m/s,

*v*= 0

⇒ t=10/3 sec

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