A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 7.0 m/s2; after 3.8 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 73.3 m/s overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch up with the other car

The distance traveled by both cars once they are in the main speedway, assuming that the acceleration of the refueling car is constant, is given by the following kinematic equation:

[tex]x = v_{o}*t + \frac{1}{2} * a * t^{2} (1)[/tex]

The refueling car (which we will call car 1) in the moment that enters to the main speedway, has achieved a speed that can be found from the definition of acceleration, rearranging terms, as follows:

[tex]v_{f1} = a* t = 7.0m/s2*3.8s = 26.6 m/s (2)[/tex]

[tex]x_{1} = v_{f1}*t + \frac{1}{2} * a * t^{2} (3)[/tex]

Now, for the other car (which we will call car 2), due to is moving at a constant speed, a=0, so we can write the following equation for x₂:

[tex]x_{2} = v_{f2}*t = 73.3m/s*t (4)[/tex]

When the entering car catches up the other car, both distances will be equal each other, so x₁ = x₂, as follows:
[tex]26.6m/s*t + \frac{1}{2} * 7.0m/s2* t^{2} = 73.3m/s*t[/tex]
Rearranging, simplifying and solving for t:
[tex]t =\frac{2*(73.3m/s-26.6m/s}{7.0m/s2} = 13.3 s (5)[/tex]