Max distance for person to catch up to bus

[circuscircus]

Homework Statement

THe bus is D=10m ahead of the person and accelerates from rest at a=2m/s^2 while the person runs after it at 7m/s. Suppose D is not specified, what maximum distance can the bus be ahead of the person that will allow the person to catch the bus?

Homework Equations

The Attempt at a Solution

I'm not really sure how to approach this...

 

[mgb_phys]

You need an equation describing the distance the bus has moved with time, and a different equation for the person.
When they catch the bus they have the same distance so set the equations equal,
look in the sticky thread for equations of motion or in your textbook.

 

[m2003]

I would approach this problem using the equations of motion and the concept of relative velocity. The distance (D) between the person and the bus can be represented as a function of time (t) using the equation D = ut + 1/2at^2, where u is the initial velocity (in this case, 7m/s) and a is the acceleration (2m/s^2).

To find the maximum distance that the bus can be ahead of the person and still allow them to catch up, we can set the final velocity of the person (v = 7m/s) equal to the final velocity of the bus (u = 0m/s) and solve for t. This will give us the time it takes for the person to catch up to the bus.

7m/s = 0m/s + 2m/s^2(t)
t = 3.5 seconds

From here, we can plug this value for t into the equation for distance and solve for D.

D = (7m/s)(3.5s) + 1/2(2m/s^2)(3.5s)^2
D = 24.5m

Therefore, the maximum distance that the bus can be ahead of the person and still allow them to catch up is 24.5m. Any distance greater than this would result in the person not being able to catch the bus, assuming they continue running at a constant speed of 7m/s.