A race driver has made a pit stop to refuel. After refueling, he leaves the pit area with an acceleration whose magnitude is 5.40 m/s2; after 4.0 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant speed of 69 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?
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To solve this problem, we need to find the time it takes for the entering car to catch up with the other car.
Let's break down the problem and find the distance traveled by each car after 4.0 seconds, when the entering car enters the main speedway.
The distance covered by the entering car after 4.0 s can be calculated using the equation:
distance = initial velocity * time + (1/2) * acceleration * (time^2)
Given:
Initial velocity of entering car = 0 m/s (since it starts from rest in the pit area)
Time = 4.0 s
Acceleration = 5.40 m/s^2
Plugging these values into the equation, we get:
distance = 0 * 4.0 + (1/2) * 5.40 * (4.0^2) = 43.2 m
Now, since the other car is already ahead of the entering car on the speedway, we need to find the distance between the two cars.
The distance traveled by the other car during the 4.0 s can be calculated using the equation:
distance = speed * time
Given:
Speed of the other car = 69 m/s
Time = 4.0 s
Plugging these values into the equation, we get:
distance = 69 * 4.0 = 276 m
So, the distance between the two cars when the entering car enters the main speedway is 276 m - 43.2 m = 232.8 m.
Now, since the entering car is accelerating and the other car is traveling at a constant speed, we can set up an equation to find the time it takes for the entering car to catch up with the other car.
Let's call the time it takes for the entering car to catch the other car as 't'.
The distance traveled by the entering car during time 't' can be calculated using the equation:
distance = initial velocity * time + (1/2) * acceleration * (time^2)
The distance traveled by the other car during time 't' can be calculated using the equation:
distance = speed * time
Since both cars are at the same position when the entering car enters the main speedway, we can set these two distances equal to each other and solve for 't'.
Initial velocity of entering car = 0 m/s (since it starts from rest at the pit stop)
Acceleration of entering car = 5.40 m/s^2
Speed of the other car = 69 m/s
So, the equation becomes:
0 * t + (1/2) * 5.40 * (t^2) = 69 * t
We can solve this equation to find 't', which represents the time it takes for the entering car to catch the other car.
0.5 * 5.40 * (t^2) = 69 * t
2.7 * (t^2) = 69 * t
2.7 * (t^2) - 69 * t = 0
Factoring out 't', we get:
t * (2.7 * t - 69) = 0
So, either t = 0, or (2.7 * t - 69) = 0
Ignoring t = 0, we solve the second equation:
2.7 * t - 69 = 0
2.7 * t = 69
t = 69 / 2.7
Calculating this value, we find:
t ≈ 25.556
Therefore, it takes approximately 25.556 seconds for the entering car to catch the other car.