# Velocity help

1. Mar 1, 2004

### boxelderbug

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 in to get the distance.

2. Mar 1, 2004

### boxelderbug

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

3. Mar 1, 2004

### NateTG

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 $$6 \frac{ft}{s^2}$$

4. Mar 3, 2004

### boxelderbug

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.