Question

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

Answer 1

t = 13.3 s

Step-by-step 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:

`x = v_(o)*t + (1)/(2) * a * t^(2) (1)`

• 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:

`v_(f1) = a* t = 7.0m/s2*3.8s = 26.6 m/s (2)`

• So, since vf1 = v₀ in (1), we get:

`x_(1) = v_(f1)*t + (1)/(2) * a * t^(2) (3)`

• 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:

`x_(2) = v_(f2)*t = 73.3m/s*t (4)`

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