Solving the Physics Problem: Car and Truck Acceleration

AI Thread Summary
The discussion revolves around a physics problem involving a car and a truck accelerating at different rates. The car starts at a constant velocity of 6.0 ft/s, while the truck travels at a constant speed of 34 ft/s. The main confusion lies in understanding the concept of acceleration, as the car's velocity is constant initially, meaning its acceleration is zero. To determine when the car will overtake the truck, the user needs to set up equations for both vehicles' positions and solve for the time and distance. Ultimately, the user realizes that taking the derivative helps clarify the acceleration concept, leading to a solution.
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I'm not really good at physics so relevant any feedback would be helpful. Here's the question;
At the instant a traffic light turns green, a car starts with a constant velocity of 6.0 ft/s. At the same instant a truck, traveling with a constant speed of 34 ft/s, overtakes and passes the car. How far beyond the traffic light will the car overtake the truck?
This is what I got so far;
The acceleration of both the truck and the car is zero.
truck: x=34(ft/s)t
Car: x=(1/2)(6.0ft/s+34ft/s)t=20t
This is where I get stuck. Can you tell me what I'm doing wrong or what I'm not doing?
I know that I will have to use two equations with two unknowns. By doing this I should get me the time it takes for the car to catch up to truck. Then I can plug the time back into get the distance.
 
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I got confused on the acceleration of the car. If an object is changing its velocity -whether by a constant amount or a varying amount - then it is an accelerating object. And an object with a constant velocity is not accelerating. Therefore, the acceleration of the car would be 28(ft/s)/t
 
You should be able to use the equation:
x(t)=x_0+v_0t+\frac{1}{2}at^2
to solve this. To see when the truck overtakes the car you can determine the time at which the two are in the same position.

BTW: I'm guessing that the acceleration of the car is 6 \frac{ft}{s^2}
 
Well thank you, I've figured it out and I'm assuming that I just had to take the derivative to get the acceleration, duh. Thanks for all of your help.
 
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