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 6.3 m/s2; after 3.6 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 68.4 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?
V1 = 68.4 m/s
X1 = 68.4 t
V2 = a (t-3.6)
X2 = (1/2) a (t-3.6)^2,
where a = 6.3 m/s^2
Set X1 = X2 and solve for t.
can you give me the answer?
To solve this problem, we will need to find the time it takes for the entering car to catch up with the other car.
Let's break down the information given:
Acceleration of the entering car, a = 6.3 m/s^2
Time taken for the entering car to enter the main speedway, t = 3.6 s
Velocity of the other car, v = 68.4 m/s
First, we need to find the distance covered by the entering car from the pit area to the main speedway. We can do this by using the equation:
s = ut + (1/2)at^2
where s is the distance covered, u is the initial velocity (which is 0 m/s because the entering car starts from rest), and t is the time interval.
Substituting the given values, we have:
s = 0 + (1/2)(6.3)(3.6^2)
Next, we need to find the time it takes for the entering car to catch up with the other car. Since the other car is already traveling at a constant velocity of 68.4 m/s, we can use the equation:
s = v(t + t')
where s is the distance covered, v is the velocity of the other car, t is the time interval from the pit area to the main speedway, and t' is the time it takes for the entering car to catch up with the other car.
Substituting the values of s and v, we have:
(1/2)(6.3)(3.6^2) = 68.4(t + t')
Now, we can solve for t' by rearranging the equation:
t' = [(1/2)(6.3)(3.6^2) - 68.4t] / 68.4
Finally, substitute the value of t given in the problem (t = 3.6s) to find the time it takes for the entering car to catch up with the other car.