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 5.8 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 67.4 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?
how fast is he going after 3.6 seconds
at = 5.8 * 3.6 = 20.88
so
starting new t = 0 at 20.88 m/s
v = 20.88 + 5.8 t
d = 20.88 t + (5.8/2) t^2
same distance as the other car going 67.4
so
67.4 t = 20.88 t + (5.8/2)t^2
solve quadratic, use reasonable of two answers.
67
To solve this problem, we need to break it down into two parts:
1. Find the initial speed of the entering car when it leaves the pitstop and enters the main speedway.
2. Determine the time it takes for the entering car to catch up with the other car.
Let's start by calculating the initial speed of the entering car.
1. Find the initial speed of the entering car:
Given:
Acceleration (a) = 5.8 m/s^2
Time (t) = 3.6 s
We can use the acceleration formula, which relates acceleration, time, and change in velocity:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the car starts from rest, the initial velocity (u) is 0. So the formula becomes:
v = 0 + at
Substituting in the given values:
v = 0 + (5.8 m/s^2)(3.6 s)
v = 20.88 m/s
Therefore, the initial speed of the entering car is 20.88 m/s when it enters the main speedway.
2. Determine the time it takes for the entering car to catch up with the other car:
Given:
Velocity of the other car (Vo) = 67.4 m/s
To find the time it takes for the entering car to catch up with the other car, we can use the equation:
d = Vo*t
where:
d = distance traveled by both cars when they meet
Vo = velocity of the other car
t = time
Since both cars meet when the entering car catches up with the other car, the distance traveled by both cars is the same.
We can also use the equation of motion for the entering car:
d = ut + 0.5*a*t^2
where:
d = distance traveled by both cars when they meet
u = initial velocity of the entering car
a = acceleration of the entering car
t = time it takes for the entering car to catch up
Using the equations for both cars, we have:
ut + 0.5*a*t^2 = Vo*t
Rearranging the equation, we get a quadratic equation in terms of t:
0.5*a*t^2 + (u - Vo)*t = 0
Now we can substitute the values:
0.5*(5.8 m/s^2)*t^2 + (20.88 m/s - 67.4 m/s)*t = 0
Simplifying the equation:
2.9*t^2 - 46.52*t = 0
To solve this quadratic equation, we can factor out t:
t * (2.9t - 46.52) = 0
So two solutions are possible:
1. t = 0 (which is not possible since the cars have to meet)
2. 2.9t - 46.52 = 0
Solving for t:
2.9t = 46.52
t = 46.52 / 2.9
t ≈ 16.07 s
Therefore, it takes approximately 16.07 seconds for the entering car to catch up with the other car.