Question

Physics help!!! How long will it take for the second-place cyclist to catch up with the leader??

The leader of a bicycle race is traveling with a constant velocity of +14.50 m/s and is 18.9 m ahead of the second-place cyclist. The second-place cyclist has a velocity of +9.20 m/s and an acceleration of +1.30 m/s2. How much time elapses before he catches the leader?

Answer 1

There will be a new leader in 10.83 s.

Explanation:

Notation: Let leader's data have subscript 1 and the data of the rider in 2nd place have subscript 2.

From the initial conditions, time will go forward at the same rate for both riders, so time will just be t.

When `"rider"_2` catches `"rider"_1`, the distance ridden by `"rider"_1` will be

`s_1 = +14.50 m/s*t`

The distance ridden by `"rider"_2` will be

`s_2 = 18.9 m +14.50 m/s*t`

This equation will describe how that is accomplished:

`s_2 = u_2*t + 1/2*a_2*t^2`
`18.9 m +14.50 m/s*t = 9.20 m/s*t + 1/2*1.30 m/s^2*t^2`
`0 = 0.65 m/s^2*t^2 + (9.20-14.50) m/s*t - 18.9 m`

I will drop the units from this point forward. It is clear that t will be in units of seconds.
`0.65 *t^2 - 5.30 *t - 18.9 = 0`

So we have a quadratic formula.

`t = (5.30 +- sqrt((-5.30)^2 -4*0.65*(-18.9)))/(2*0.65)`

`t = (5.30 +- sqrt(28.09 + 49.14))/(2*0.65)`

`t = (5.30 +- sqrt(77.23))/(1.30)`

`t = (5.30 +- 8.788)/(1.30)`
One of the solutions will be negative time, so that one can be ignored.

`t = (5.30 + 8.788)/(1.30) = 14.088/1.30 = 10.83 s`

I hope this helps,
Steve