A truck is stopped for a red light at Ridge Road on Higgins line. At the instant the light turns green, the truck pulls away with a constant acceleration of 3.3 m/s2. A cyclist approaching the truck with a constant velocity of 12.0 m/s is 8.60 m behind the truck when the light turns green.
(1) What is the elapsed time from the changing of the light to when the cyclist passes the truck?
(2)What is the elapsed time from the changing of the light to when the truck repasses the cyclist?
i know how to do number 1. but i'm stuck on number two. i have the answer (6.38s) but i have no idea how to get that answer. any help would be greatly appreciated! thanks
To solve for the elapsed time from the changing of the light to when the truck repasses the cyclist, we can start by finding the position of the truck and cyclist at the moment they pass each other.
1. We know that the truck has a constant acceleration of 3.3 m/s^2 from the moment the light turns green. We can use the equation of motion: s = ut + 1/2at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is time.
The initial velocity of the truck is 0 m/s since it starts from rest. Thus, the displacement of the truck at the moment the cyclist passes it is given by:
s_truck = 1/2 * 3.3 * t^2 (equation 1)
2. The cyclist, on the other hand, has a constant velocity of 12.0 m/s. The displacement of the cyclist at the moment of passing the truck is given by:
s_cyclist = 12.0 * t (equation 2)
Now, let's find the time when the cyclist passes the truck.
3. Since they pass each other, when the cyclist overtakes the truck, their respective displacements will be equal. Therefore, we can equate equations 1 and 2:
1/2 * 3.3 * t^2 = 12.0 * t
4. Simplify the equation:
1.65 * t^2 = 12.0 * t
5. Rearrange the equation to form a quadratic equation:
1.65 * t^2 - 12.0 * t = 0
We can now solve this equation using the quadratic formula.
6. Apply the quadratic formula to find the positive value of t:
t = (-b + √(b^2 - 4ac)) / (2a), where a = 1.65, b = -12.0, and c = 0
t = (-(-12.0) + √((-12.0)^2 - 4 * 1.65 * 0)) / (2 * 1.65)
t = (12.0 + √(144.0 - 0)) / 3.3
t = (12.0 + √144.0) / 3.3
t ≈ (12.0 + 12.0) / 3.3
t ≈ 24.0 / 3.3
t ≈ 7.27 seconds
The time it takes for the cyclist to pass the truck is approximately 7.27 seconds.
Now, let's move on to solving the second part of the question: the elapsed time from the changing of the light to when the truck repasses the cyclist.
7. To find the elapsed time from the changing of the light to when the truck repasses the cyclist, we need to find the time it takes for the truck to catch up to the cyclist after they pass each other.
8. As we found earlier, the time it takes for the cyclist to pass the truck is approximately 7.27 seconds.
9. Since the truck starts from rest and accelerates at 3.3 m/s^2, we can again use the equation of motion: s = ut + 1/2at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is time.
Let's denote the time it takes for the truck to repass the cyclist as T. At this moment, the displacement of the truck is equal to the displacement of the cyclist.
Using equation 1, we can write:
1/2 * 3.3 * T^2 = 12.0 * T
10. Simplify the equation:
1.65 * T^2 - 12.0 * T = 0
11. Rearrange the equation to form a quadratic equation:
1.65 * T^2 - 12.0 * T = 0
Now, solve this quadratic equation using the quadratic formula to find the positive value of T:
T = (-(-12.0) + √((-12.0)^2 - 4 * 1.65 * 0)) / (2 * 1.65)
T = (12.0 + √(144.0 - 0)) / 3.3
T = (12.0 + √144.0) / 3.3
T ≈ (12.0 + 12.0) / 3.3
T ≈ 24.0 / 3.3
T ≈ 7.27 seconds
Therefore, the elapsed time from the changing of the light to when the truck repasses the cyclist is approximately 7.27 seconds, which is indeed the same as the time it takes for the cyclist to pass the truck.