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 5.4 m/s2; after 3.7 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 68.0 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?
To solve this problem, we need to determine the time it takes for the entering car to catch up with the other car. Let's break down the problem step by step:
1. First, we need to find the initial velocity of the entering car when it leaves the pit area. Since it starts from rest and accelerates at a rate of 5.4 m/s² for 3.7 s, we can use the following formula to find the final velocity:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Plugging in the values, we have:
v = 0 + (5.4 m/s²)(3.7 s)
v = 19.98 m/s (rounded to two decimal places)
Therefore, the initial velocity of the entering car is 19.98 m/s.
2. We know that the other car is traveling at a constant velocity of 68.0 m/s. To catch up with the other car, the entering car needs to cover the distance between them. Let's call this distance D.
3. Now, we can calculate the time it takes for the entering car to catch up with the other car. We'll assume that both cars maintain their respective velocities and accelerations throughout this time.
We'll use the formula:
D = ut + 0.5at²
where D is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Rearranging the equation to solve for t, we have:
t = (-u ± sqrt(u² + 2aD)) / a
Since the entering car is trying to catch up to the other car, D will be negative, representing the distance covered in the negative direction.
Plugging in the values, we have:
t = (-(19.98 m/s) ± sqrt((19.98 m/s)² + 2(5.4 m/s²)(-D))) / (5.4 m/s²)
4. We know that at the moment the entering car enters the main speedway, the other car overtakes and passes it. Therefore, the distance D is the sum of the distances covered by both cars during the 3.7 s.
Distance covered by the entering car = (initial velocity) * (time) + 0.5 * (acceleration) * (time)²
Distance covered by the entering car = (19.98 m/s) * (3.7 s) + 0.5 * (5.4 m/s²) * (3.7 s)²
Distance covered by the other car = (constant velocity) * (time)
Distance covered by the other car = (68.0 m/s) * (3.7 s)
Therefore, the total distance covered by both cars is:
D = (19.98 m/s) * (3.7 s) + 0.5 * (5.4 m/s²) * (3.7 s)² - (68.0 m/s) * (3.7 s)
5. Plug the value of D into the equation for t:
t = (-(19.98 m/s) ± sqrt((19.98 m/s)² + 2(5.4 m/s²)(-D))) / (5.4 m/s²)
Calculate the value of t using this equation, and choose the positive root because time cannot be negative.
This will give you the time required for the entering car to catch up with the other car.