Question

A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s

Answer 1

The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
$S= \frac{1}{2} a t^2$
where a is the acceleration. Using the data of the problem, we can find a:
$a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2$
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J$

and so the power output is
$P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W$

b) power output at t=4.0s 
The velocity at t=4.0 s is
$v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J$

and so the power output is
$P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W$

c) Power output at t=6.0 s
The velocity at t=2.0 s is
$v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s$

the kinetic energy of the sprinter is
$K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J$

and so the power output is
$P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W$