Solving the Problem: Calculating Distance After the Start

[rayj098]

1. The problem statement

At the instant when the traffic light turns green, a car starts with a constant acceleration of 1.8 m/s^2 [forward]. At the same instant a truck traveling with a constant velocity of 8.5 m/s [forward] overtakes and passes the car. How far from the starting point will the car catch up to the truck?



2. The attempt at a solution
Car:
d = V1*t + 1/2*a*t^2
d= 0 + 1/2(1.8)(t)^2
d= 0.9t^2

Truck:
d = V * t
d = 25t


Put it all together

0.9t^2 = 25t
0.9t^2 - 25t = 0


Now what?

 

[whoareyou]

I think your equation for your truck is wrong. After you have the two equations, put them together and solve for time.

 

[rayj098]

How is the truck equation wrong? Please expand

 

[whoareyou]

In the question, it says that the truck is traveling at a constant velocity of 8.5m/s [F] and in your equation, you have 25 as the velocity.

 

[rwisz]

To solve this problem, we can use the quadratic formula to find the time at which the car catches up to the truck. The formula is t = (-b ± √(b^2 - 4ac)) / 2a, where a = 0.9, b = -25, and c = 0. Plugging these values in, we get t = (-(-25) ± √((-25)^2 - 4(0.9)(0))) / 2(0.9) = (25 ± √(625)) / 1.8 = (25 ± 25) / 1.8 = 50 / 1.8 or 0 / 1.8. This gives us two solutions, t = 27.78 seconds or t = 0 seconds.

Since we are looking for the time at which the car catches up to the truck, we can discard the solution t = 0 seconds, as the car and truck are already at the same position at the start. Therefore, the car will catch up to the truck at t = 27.78 seconds.

To find the distance from the starting point, we can plug this time back into either of the equations for the car or the truck. Using the equation for the car, d = 0.9(27.78)^2 = 680.4 meters. Therefore, the car will catch up to the truck at a distance of 680.4 meters from the starting point.