Answer:
The car will pass the truck 25 s after leaving the traffic light
Explanation:
These equations apply for a straight movement with constant acceleration:
x = x0 + v0 t + 1/2 a t²
v = v0 + a t
where:
x = position at time t
x0 = initial position
a = acceleration
v0 = initial speed
v = speed
t = time
For a straight movement with constant speed, as the truck:
x = x0 + v t
First, let´s calculate how much time the car accelerates, using the equation of speed:
v = v0 + a t
25 m/s = 0 m/s + 1.25 m/s² * t
t = 25 m/s / 1.25 m/s²
t = 20 s
Now, let´s see the position of the car and the truck at that time:
The position of the car will be:
x = x0 + v0 t + 1/2 a t²
since it starts from rest and considering the the traffic light the initial point:
v0 = 0
x0 = 0
then
x = 1/2 *1.25 m/s² * (20 s)² = 250 m
The truck position at that time will be:
x = x0 + v * t = 15 m/s * 20 s = 300 m (considering the traffic light as the initial position)
Then, the truck is 50 m ahead of the car when the car reaches it´s maximum speed.
Now, we have to find the time at which the car and the truck have the same position.
For the car, the equation for the position is:
x = x0 + v t
Now, let´s consider x0 as the position at which the car starts traveling at constant speed, that is, 250 m.
For the truck, the equation of the position will also be:
x = x0 + v t
And now x0 will be the position at t = 20 s, when the car starts traveling at constant speed, that is, 300 m. So, we have a chase in which the truck starts at the same time that the car but 50 m ahead of it traveling at 15 m/s while the car travels at 25 m/s.
Since the position of the car and the truck will be the same when the car passes the truck:
x car = x truck
x0 car + v car * t = x0 truck + v truck * t
250 m + 25 m/s * t = 300 m + 15 m/s * t
25 m/s * t - 15 m/s * t = 300 m - 250 m
10 m/s * t = 50 m
t = 50 m / 10 m/s = 5 s
After 25 s, the car will pass the truck.
