Question

A 62.0 kg sprinter starts a race with an acceleration of 1.44 m/s2. If the sprinter accelerates at that rate for 30 m, and then maintains that velocity for the remainder of the 100 m dash, what will be his time (in s) for the race

Answer 1

Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

$v^{2}-u^{2}= 2as$

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$.

$v^{2}-(0)^{2}= 2(1.44)(30)$

$v= 9.3 m^{-1}$

Calculate time taken to cover 30 m distance.

The expression for the equation of motion is as follows;

$s=ut+\frac{(1)}{2}at^2$

Put u= 0, s=30 m and $a= 1.44 ms^{-2}$.

$30=(0)t+\frac{(1)}{2}(1.44)t^2$

t=6.45 s

Calculate the time taken to complete his race.

T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

$T=6.45+\frac{s'}{v}$

Put s= 30 m, $v= 9.3 m^{-1}$ and s'= 100 m.

$T=6.45+\frac{(100)}{9.3}$

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.