Integrated Concepts When starting a foot race, a 70.0-kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s.

(a) What is his final speed?

(b) How far does he travel?

• Mass of the sprinter = 70 kg.
• Time the force exerted = 0.800 s.
• Force exerted = 650 N.

Apply Newton’s Second Law of motion as:

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}$

Here, m is the mass of the sprinter, a is the acceleration, and F is the force exerted.

Substitute 70 kg for m and 650 N for F in the above expression, and we get,

$\begin{array}{c}\text{650\hspace{0.33em}N}=\text{70\hspace{0.33em}kg}\times a\\ a=\frac{650\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^2}{70\text{\hspace{0.33em}kg}}\\ a=9.286{\text{\hspace{0.33em}m/s}}^2\end{array}$

Apply the equation of motion as:

$\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$

Here, v is the final speed, u is the initial speed, and t is the time taken.

Substitute 0 m/s for u, 9.286 m/s²for a, and 0.8 s fortin the above equation, and we get,

$\begin{array}{c}v=0+9.286{\text{\hspace{0.33em}m/s}}^2\times 0.8\text{\hspace{0.33em}s}\\ =7.43\text{\hspace{0.33em}m/s}\end{array}$

Hence, the final speed of the sprinter is 7.43 m/s.

Apply the equation of motion as:

$\mathit{s}\mathbf{=}\mathit{u}\mathit{t}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$

Here, s is the distance traveled.

Substitute 0 m/s for u, 9.286 m/s²for a, and 0.8 s for t.

$\begin{array}{c}s=0+\frac{1}{2}\times 9.286{\text{\hspace{0.33em}m/s}}^{\text{2}}\times {0.8}^2{\text{\hspace{0.33em}s}}^{\text{2}}\\ =\frac{1}{2}\times 9.286{\text{\hspace{0.33em}m/s}}^{\text{2}}\times 0.64{\text{\hspace{0.33em}s}}^{\text{2}}\\ =\frac{1}{2}\times 5.94\text{\hspace{0.33em}m}\\ =2.97\text{\hspace{0.33em}m}\end{array}$

Hence, the distance traveled by the sprinter is 2.97 m.