What is the runner's speed?

Physics Problem 

A cyclist from a state of rest begins a rectilinear, uniformly accelerated movement along a bicycle path at the moment when a person runs past him at a constant velocity. The speed of the cyclist at the moment when he caught up with the runner is Vc = 8 m/s. What is the runner's speed Vr in m/s at this moment? Input numerical part of your answer.

Solution

Let's denote the time it takes for the cyclist to catch up with the runner as

$�$ seconds. Since the cyclist starts from rest and moves with uniform acceleration, we can use the equation for the final velocity of an object undergoing uniform acceleration:

${�}_{�}=�\cdot �$

where ${�}_{�}$ is the final velocity of the cyclist (8 m/s) and $�$ is the acceleration of the cyclist. Rearranging the equation, we get:

$�=\frac{{�}_{�}}{�}$

Now, the distance covered by the cyclist and the runner is the same when the cyclist catches up with the runner. For the cyclist, the distance $�$ can be calculated using the equation for distance covered under uniform acceleration:

$�=\frac{1}{2}�{�}^2$

Substituting the expression for $�$ from above:

$�=\frac{1}{2}\left(\frac{{�}_{�}}{�}\right){�}^2=\frac{1}{2}{�}_{�}�$

For the runner, who is moving at a constant velocity ${�}_{�}$, the distance covered is simply:

$�={�}_{�}\cdot �$

Equating the two distances:

$\frac{1}{2}{�}_{�}�={�}_{�}\cdot �$

Solving for ${�}_{�}$:

${�}_{�}=\frac{1}{2}{�}_{�}$

Substituting the value of ${�}_{�}$ (8 m/s):

${�}_{�}=\frac{1}{2}\times 8=4\text{ m/s}$

So, the speed of the runner at the moment when the cyclist catches up with him is 4 m/s.