A race driver has made a pit stop to refuel. After refueling, he leaves the pit area with an acceleration whose magnitude is 6.0m/s 2 . After 4.0s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant speed of 70 m/s overtakes and passes the entering car. If the entering car maintains its acceleration, how much time is required for it to catch the other car?

Measure time from the instant the pit car (#1) enters the speedway. When that car enters, its speed is 6.0 * 4.0 = 24 m/s, and it is still accelerating.

X1 = Vo t + (1/2) a t^2 = 24t + 3 t^2
For the constant-speed car,
X2 = 70 t
The time required for car 1 to catch up with car #2 is given by
X1 = X2
24t + 3 t^2 = 70t
46t = 3 t^2
t = 0 or 46/3 s
We are not interested in the t=0 solution, therefore
t = 15.3 s

Why did the entering car go to the same school as the other car? Because they wanted to "catch" up on their education! But it's not all bad for the entering car. With an acceleration of 6.0 m/s^2, it's got some serious catching up to do! Now, let's calculate how much time it will take for the entering car to catch the other car.

To solve this, we can use the kinematic equation:

v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time

The entering car's initial velocity is 0 m/s (since it just left the pit stop), and we want to find the time it takes for it to catch up with the other car, so we'll call that time t.

The other car's speed is 70 m/s, and we know that it takes 4.0 seconds for the entering car to enter the main speedway. So now, let's find the entering car's final velocity at that 4.0-second mark.

v = u + at
v = 0 + 6.0 * 4.0
v = 24.0 m/s

Now we've got both cars traveling at the same speed, 24.0 m/s. But we want to know how much time it takes for the entering car to catch up with the other car, which is still 70 m/s.

To catch up to the other car, the entering car needs to cover the difference in their speeds, which is (70 - 24) = 46 m/s.

Now we can use our formula again, this time with the final velocity as 70 m/s:

v = u + at
70 = 24 + 6 * t

Rearranging the equation to solve for t:

6t = 70 - 24
6t = 46
t = 46 / 6
t ≈ 7.67 seconds

So, it will take approximately 7.67 seconds for the entering car to catch up with the other car. I hope they brought some snacks for the road trip!

To solve this problem, we need to analyze the motion of both cars and find the time it takes for the entering car to catch the other car.

Let's start by determining the position of the entering car when it enters the main speedway. We can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = displacement (position)
u = initial velocity (0 m/s)
a = acceleration (6.0 m/s^2)
t = time (4.0 s)

Plugging in the values, we have:

s = (0)(4.0) + (1/2)(6.0)(4.0^2)
s = 0 + (1/2)(6.0)(16.0)
s = 0 + (1/2)(96.0)
s = 0 + 48.0
s = 48.0 m

Therefore, the entering car is 48.0 meters behind the other car when it enters the main speedway.

Now, let's calculate the relative speed between the two cars:

relative speed = speed of the overtaking car - speed of the entering car
relative speed = 70.0 m/s - 0 m/s
relative speed = 70.0 m/s

To catch up to the other car, the entering car needs to cover the 48.0 meters distance between them at a relative speed of 70.0 m/s.

We can use the equation of motion again to find the time it takes:

s = ut + (1/2)at^2

Where:
s = distance (48.0 m)
u = initial velocity (0 m/s)
a = acceleration (6.0 m/s^2)
t = time (unknown)

Plugging in the values, we have:

48.0 = (0)(t) + (1/2)(6.0)(t^2)
48.0 = 0 + (1/2)(6.0)(t^2)
48.0 = 3.0t^2

Divide both sides by 3.0:

16.0 = t^2

Taking the square root of both sides:

t = √16.0
t = 4.0 s

Therefore, it will take the entering car 4.0 seconds to catch the other car.

To solve this problem, we need to analyze the motion of both cars and find the time it takes for the entering car to catch up to the other car.

Let's break down the problem step by step:

Step 1: Determine the initial velocity and acceleration of the entering car.
- The problem states that after leaving the pit area, the entering car has an acceleration of 6.0 m/s^2. However, it does not mention the initial velocity of the car. We need to find the initial velocity from the given information.

Step 2: Calculate the distance covered by the entering car before entering the main speedway.
- The entering car spends 4.0 seconds accelerating before entering the main speedway. To calculate the distance covered during this time, we need to use the following kinematic equation:
s = ut + (1/2)at^2
where:
- s is the distance covered (unknown)
- u is the initial velocity (unknown)
- a is the acceleration (6.0 m/s^2)
- t is the time (4.0 seconds).

Step 3: Calculate the distance covered by the entering car on the main speedway.
- The entering car starts from zero velocity when it enters the main speedway. We need to find out how much distance it covers before catching up to the other car.
- The other car has a constant speed of 70 m/s, and the entering car's motion is accelerated. To catch up, the entering car needs to cover the same distance as the other car.

Step 4: Calculate the time required for the entering car to catch up to the other car.
- We will use the equation of motion for the entering car, where the final distance covered is equal to the distance covered by the other car (from step 3).
s = ut + (1/2)at^2
where:
- s is the distance covered by the entering car on the main speedway (from step 3)
- u is the initial velocity of the entering car (from step 1)
- a is the acceleration of the entering car (6.0 m/s^2)
- t is the time we need to calculate.

By following these steps, we can find the time it takes for the entering car to catch up to the other car.