A truck is stopped at a stoplight. When the light turns green, the truck accelerates at 2.5 meters per second squared. At the same instant, a car passes the truck going 15 meters per second. Where and when does the truck catch up with the car?
car's distance travelled X1 = 15 t
truck's distance travelled X2 = (1/2)2.5 t^2 = 1.25 t^2
Set X1 =X2 and solve for t. We are looking for the solution that is not t=0.
1.25 t^2 = 15 t
t^2 - 12 t = 0
t (t-12) = 0
To find the location and time when the truck catches up with the car, we can use kinematic equations.
Let's assign variables:
- Let "t" be the time it takes for the truck to catch up with the car.
- Let "d" be the distance between the starting point and the catch-up point.
- Let "v_c" be the constant velocity of the car.
- Let "a_t" be the constant acceleration of the truck.
The equation for the distance covered by the car, d_car, is given by:
d_car = v_c * t
The equation for the distance covered by the truck, d_truck, is given by:
d_truck = 0.5 * a_t * t^2
Since the truck starts from rest, its initial velocity is 0.
For the truck to catch up with the car, the distances covered by both should be equal. So we have:
0.5 * a_t * t^2 = v_c * t
Solving this equation will give us the time it takes for the truck to catch up with the car.
Now, let's calculate it:
0.5 * (2.5) * t^2 = 15 * t
Rearranging the equation:
1.25 * t^2 = 15 * t
Dividing both sides by "t" (assuming t ≠ 0):
1.25 * t = 15
Solving for "t":
t = 15 / 1.25
t = 12 seconds
So it will take 12 seconds for the truck to catch up with the car.
To find the location where the catch-up occurs, we can substitute this time back into either of the distance equations:
d_truck = 0.5 * a_t * t^2
d_truck = 0.5 * (2.5) * (12)^2
d_truck = 180 meters
Therefore, the truck will catch up with the car after 12 seconds at a location 180 meters from the starting point.