# The average marathon runner can complete the 42.2-km distance of the marathon in 3 h and 30 min....

## Question:

The average marathon runner can complete the 42.2-km distance of the marathon in 3 h and 30 min. If the runner's mass is 75 kg, what is the runner's average kinetic energy during the run?

## Kinetic Energy:

{eq}\\ {/eq}

Kinetic Energy is one of the two types of mechanical energy present in an object (the other being the potential energy). This energy is solely due to the motion of the object.

Formally, it is defined as the energy needed to accelerate an object from rest to its current velocity. The kinetic energy of an object:

• Increases with the mass of the object.
• Increases with the velocity of the object.

The kinetic energy of an object does not depen on how the object was accelerated.

{eq}\\ {/eq}

We are given:

• The mass of the runner, {eq}m=75\;\rm kg {/eq}
• The total distance covered by the runner, {eq}\Delta s=4.2\;\rm km=4.22\times 10^{4}\;\rm m {/eq}
• The total time taken by the runner, {eq}\Delta t=3\;\rm h,\,30\;\rm min=3\times 3600+30\times 60\;\rm s=12600\;\rm s {/eq}

The kinetic energy of an object with mass, {eq}m {/eq}, moving with velocity, {eq}v {/eq}, is given by the equation:

$$K=\dfrac{1}{2}mv^2$$

The average kinetic energy is obtained by considering the average velocity of the object.

The average velocity of an object is given by the following equation:

$$v=\dfrac{\Delta s}{\Delta t}$$

Here,

• {eq}\Delta s {/eq} is the displacement of the object in time {eq}t {/eq}

After plugging the given values into the above equation, we have:

{eq}\begin{align*} v&=\dfrac{4.22\times 10^{4}}{1.26\times 10^{4}}\\ &\approx 3.35\;\rm m/s \end{align*} {/eq}

Therefore, the average kinetic energy of the runner is:

{eq}\begin{align*} K&=\dfrac{1}{2}\times 75 \times \left (3.35 \right )^2\\ &\approx \boxed{421\;\rm J} \end{align*} {/eq}