Question

The friction coefficient between an athlete’s shoes and the ground is $0.90$ . Suppose a superman wears these shoes and races for $50\,m$ . There is no upper limit on his capacity of running at high speeds. Suppose he takes exactly this minimum time to complete the $50\,m$ , what minimum time will he take to stop?

Answer 1

Hint: Use the formula of the frictional coefficient to find the maximum acceleration. Substitute the obtained acceleration, initial velocity and the displacement in the law of the equation of motion, to find minimum time required to complete the given distance.

Formula used:
(1) The formula of the friction coefficient is
$a = \mu g$
Where $a$ is the acceleration, $\mu $ is the frictional coefficient between the shoe and the road and $g$ is the acceleration due to gravity.
(2) The equation of the law of motion is given by
$s = ut + \dfrac{1}{2}a{t^2}$
Where $s$ is the displacement of the run, $u$ is the initial velocity, $t$ is the time taken for the run and $a$ is the acceleration.

Complete step by step answer:
The friction coefficient between the shoe and the ground, $\mu = 0.90$
The displacement, $s = 50\,m$
If the minimum time is used to complete the run, the maximum acceleration is used. The maximum acceleration is calculated by using the formula (1),
$a = \mu g$
Substituting the value of the frictional coefficient and the value of the $g$ as $10$ in the above formula, we get.
$a = 0.9 \times 10$
By performing the multiplication in the above step
$a = 9\,m{s^{ - 2}}$
Hence the maximum acceleration is obtained as $9\,m{s^{ - 2}}$ .
By using the formula of the equation of the law of motion,
$s = ut + \dfrac{1}{2}a{t^2}$
Substituting the value of $u = 0$, and other known values.
$50 = 0 + \dfrac{1}{2}9{t^2}$
By simplifying the above equation, we get
$t = \sqrt {\dfrac{{100}}{9}} $
$t = \dfrac{{10}}{3}$
By again simplifying the value,
$t = 3.33\,s$

Hence, the minimum time to complete the run is $3.33\,s$.

Note: The value of the acceleration due to gravity is obtained as $9.81\,m{s^{ - 1}}$ and it is substituted in the formula. The athlete will cover maximum distance when his acceleration is maximum and hence the maximum distance can be covered in very small time.