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

t = 13.3 s

Explanation:

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:

       

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:

       

So, since vf1 = v₀ in (1), we get:

       

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:

       

When the entering car catches up the other car, both distances will be equal each other, so  x₁ = x₂, as follows:   Rearranging, simplifying and solving for t:    

Ithink your answer is d

Iwas going to ask that quesytion  sorry dude i dont  know : (    : /