A sprinter who weighs670 Nruns the first 7.0 mof a race in1.6 s, starting from rest and accelerating uniformly. What are the sprinter’s

1. Speed and
2. Kinetic energy at the end of the1.6 s?
3. What average power does the sprinter generate during the1.6 sinterval?

Distance, S = 7.0 m

Time t = 1.6 s

Force, F = 670 N

Initial velocity, $v_0=0$

Use the four equations of motion to solve such problems. Use the first and second equations of motion to get the value for acceleration and velocity accordingly.

The average power of the sprinter generated during the given time interval can be found by dividing the kinetic energy produced by the sprinter.

Formulae:

Equations of motion,

$$\begin{gathered} S=vt+\frac{1}{2}a{t}^2 \\ {v}_f=v_0+at \end{gathered}$$

Here, S is the displacement, v is the average velocity, $v_f$is the final velocity, $v_0$ is the initial velocity, a is the acceleration, and t is time.

Kinetic energy is define by,

$KE=\frac{1}{2}m{v}^2$

Here, m is the mass.

Average power is given by,

$P_{avg}=\frac{KE}{t}$

From the equation of motion, you have

$$\begin{gathered} S=v_{0}t+\frac{1}{2}a{t}^2 \\ 7.0\mathrm{m}=0+0.5\times \mathrm{a}\times {\left(1.6\mathrm{s}\right)}^2 \\ \mathrm{a}=5.469\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Using the second equation of motion,

$$\begin{gathered} \mathrm{v}={\mathrm{v}}_0+\mathrm{at} \\ =0+5.469\mathrm{m}/{\mathrm{s}}^2\times 1.6\mathrm{s} \\ =8.8\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the sprinter is 8.8 m/s .

You know that kinetic energy is,

$KE=\frac{1}{2}m{v}^2$

But from Newton’s second law,

$$\begin{gathered} \mathrm{m}=\mathrm{F}/\mathrm{g} \\ =\frac{670\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2} \\ =68.36\mathrm{kg} \end{gathered}$$

Therefore, the kinetic energy will be,

$$\begin{gathered} \mathrm{KE}=0.5\times \left(68.36\mathrm{kg}\right)\times {\left(8.8\mathrm{m}/\mathrm{s}\right)}^2 \\ =2617.18\mathrm{J} \\ =2.6\mathrm{kJ} \end{gathered}$$

Hence, the sprinter’s kinetic energy is 2.6 kJ.

Average power can be written as

$$\begin{gathered} P_{avg}=\frac{KE}{t} \\ =\frac{2617.18\mathrm{J}}{1.6s} \\ =1635.74\mathrm{W} \\ =1.6\mathrm{kW} \end{gathered}$$

Hence, the average power is 1.6 W.