a car and a truck starts from rest at the same instant with the car is behind the truck initially at the same distance. the truck has a constant acceleration of 4 ft/s^2 and the car has an acceleration of 6 ft/s^2. the car overtake the truck after the truck travelled 150 ft.
a.at what time did the car overtakes the truck?
b. how far was the car behind the truck initially?
c. what is the velocity of each when they abreast?
a. 0.5*a*t^2 = 150. 0.5*4.6t^2 = 150, t = 8.1 s.
b. 0.5a*t^2 = d + 150. a = 6 ft./s^2, t = 8.1s, d = ?.
To solve this problem, we need to use the equations of motion. Let's consider the motion of the car and truck separately.
a. To determine the time at which the car overtakes the truck, we need to find the time when their displacements are equal. We'll use the equation:
displacement = initial velocity * time + (1/2) * acceleration * time^2
For the truck:
150 ft = 0 ft/s * t + (1/2) * 4 ft/s^2 * t^2
Simplifying the equation:
2t^2 = 150 / 2
t^2 = 75 / 2
t ≈ √37.5
Therefore, the car overtakes the truck at approximately t = √37.5 seconds.
b. To determine how far the car was behind the truck initially, we'll calculate the displacement of the truck at the time the car overtakes it. Using the same equation as before, but substituting t = √37.5:
displacement_truck = 0 ft/s * √37.5 + (1/2) * 4 ft/s^2 * (√37.5)^2
= 0 ft + (1/2) * 4 ft/s^2 * 37.5 s
= 0 ft + 75 ft
= 75 ft
Therefore, the car was initially behind the truck by a distance of 75 ft.
c. At the point of overtaking, the displacements of the car and truck are equal. We can calculate the velocities of the car and truck at this point by using the equation:
final velocity = initial velocity + acceleration * time
For the car:
final velocity_car = 0 ft/s + 6 ft/s^2 * √37.5 s
For the truck:
final velocity_truck = 0 ft/s + 4 ft/s^2 * √37.5 s
So, the velocity of the car when they are abreast is approximately 6 ft/s^2 * √37.5 s, and the velocity of the truck is approximately 4 ft/s^2 * √37.5 s.