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 find the time it takes for the entering car to catch up with the other car. We can use kinematic equations to do this. Let's break down the steps:

Step 1: Find the initial velocity of the entering car.
Since the entering car starts from rest, its initial velocity is 0 m/s.

Step 2: Find the displacement of the entering car when it enters the main speedway.
To find the displacement, we can use the formula:
displacement = initial velocity * time + (1/2) * acceleration * time^2

Given:
initial velocity (u) = 0 m/s
acceleration (a) = 5.4 m/s^2
time (t) = 3.7 s

Plugging in these values, we have:
displacement = 0 * 3.7 + (1/2) * 5.4 * (3.7)^2
displacement = 0 + (1/2) * 5.4 * 13.69
displacement = (1/2) * 73.986
displacement = 36.993 m (rounded to three decimal places)

Step 3: Find the relative speed between the entering car and the other car.
The relative speed between the two cars is the difference in their velocities. Given that the other car's velocity is 68.0 m/s, the relative speed is:
relative speed = 68.0 m/s - 0 m/s
relative speed = 68.0 m/s

Step 4: Find the time taken for the entering car to catch up with the other car.
To find the time taken, we can use the formula:
time taken = displacement / relative speed

Plugging in the values, we have:
time taken = 36.993 m / 68.0 m/s
time taken ≈ 0.544 s (rounded to three decimal places)

Therefore, it takes approximately 0.544 seconds for the entering car to catch up with the other car.

To find the time it takes for the entering car to catch up with the other car, we need to determine the distance the other car travels during this time.

Let's assume the distance the entering car travels before catching up is "x" meters.

Since the entering car starts from rest and has an acceleration of 5.4 m/s^2, we can use the following kinematic equation to find the distance it travels in time t:

s = ut + 0.5at^2

Here, u is the initial velocity (0 m/s), a is the acceleration (5.4 m/s^2), and t is the time.

So, for the entering car:
s1 = 0 + 0.5 * 5.4 * (3.7^2)
s1 = 0 + 0.5 * 5.4 * 13.69
s1 = 36.846 m

Now, let's find the distance the other car travels in the same time.

The other car is traveling at a constant velocity of 68.0 m/s for 3.7 s. Therefore, its distance traveled is:

s2 = v * t
s2 = 68.0 * 3.7
s2 = 251.6 m

The entering car catches up with the other car when their distances are equal, so:

s1 = s2
36.846 = 251.6 + x
x = 36.846 - 251.6
x = -214.754 m

Since we are interested in the positive distance, we'll consider the magnitude of x:

x = 214.754 m

Now, let's find the time it takes for the entering car to catch up with the other car:

To find this time, we can use the formula:
s2 = v * t
214.754 = 68.0 * t
t = 214.754 / 68.0
t ≈ 3.16 s

Therefore, it takes approximately 3.16 seconds for the entering car to catch up with the other car.