At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.55 m/s2 . At the same instant a truck, traveling with a constant speed of 22.0 m/s , overtakes and passes the car.
a. How far beyond its starting point does the car overtake the truck?
b. How fast is the car traveling when it overtakes the truck?
1/2 a t^2 = v t ... 1/2 * 3.55 * t^2 = 22.0 t ... 1/2 * 3.55 t = 22.0
a. d = v t = 22.0 t
b. v = a t = 3.55 t
To find the answers to the questions, we need to analyze the motion of the car and the truck and determine when and where they meet.
First, let's calculate the time it takes for the car to overtake the truck.
We can use the equation of motion:
s = ut + (1/2)at^2
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
Since the car starts from rest, its initial velocity is 0 m/s. The acceleration of the car is given as 3.55 m/s^2.
Now, let's calculate the time it takes for the car to catch up with the truck:
s (car) = ut + (1/2)at^2
s (car) = 0t + (1/2)(3.55)t^2
We'll set this equation equal to the distance traveled by the truck, which is given as 22.0 m:
22.0 = (1/2)(3.55)t^2
Simplifying the equation:
44.0 = 3.55t^2
Divide both sides by 3.55:
t^2 = 44.0/3.55
t^2 ≈ 12.39
Taking the square root of both sides:
t ≈ √12.39
t ≈ 3.52 seconds
So, it takes approximately 3.52 seconds for the car to overtake the truck.
a. Now, let's find the distance the car travels during this time.
Using the equation of motion:
s (car) = ut + (1/2)at^2
s (car) = 0(3.52) + (1/2)(3.55)(3.52)^2
Simplifying the equation:
s (car) = 0 + (1/2)(3.55)(12.38)
s (car) ≈ 21.89 meters
Therefore, the car overtakes the truck approximately 21.89 meters beyond its starting point.
b. To find the speed of the car when it overtakes the truck, we can use the equation of motion once again.
v (car) = u + at
v (car) = 0 + 3.55(3.52)
v (car) ≈ 12.48 m/s
So, when the car overtakes the truck, it is traveling at approximately 12.48 m/s.