Equation for Location of Air Car on Air Track w/ Average Velocity L/t

[sfritz09]

1. L is the length of the air car, t is the time the car blocks a photogate. The car starts at a distance d from the photogate. The air track is also elevated on one end, but I don't know how much.I have to find the equation for the location of the front of an air car on an air track when the average velocity is L/t. The average speed is also L/t in the vicinity of the photogate.2.Is the average speed the same as average velocity? Except speed doesn't have direction, so it is instantaneous velocity. Would average velocity L/t be when the photogate lines up with the center of the car? 3. From what I was thinking, the location of the front of the car would be 1/2L past the photo gate, But I am supposed to express that in an equation using d, L, and t. Would the acceleration need to be found? my first guess is x=1/2(Vo+V)t...which would be x=1/2(L/t)t...which would reduce to x=1/2L. But that doesn't seem right to me. I did not use d.

[--L--]____________d_____________[___] <---Photogate
^car ^distance from photogate

 

[jbsteele]

Yes, average speed and average velocity are the same. Average velocity is the direction-specific version of average speed. The equation for the location of the front of an air car on an air track when the average velocity is L/t is x = d + 1/2L - Lt. This equation can be derived by noting that the car starts at a distance d from the photogate, and then it moves with a velocity of L/t until it reaches the photogate, covering a total distance of 1/2L (distance = rate*time). Subtracting this distance from the original distance d gives us the final location of the car's front relative to the photogate.

 

[tomcue]

I would first clarify the question and make sure all the necessary information is provided. For example, what is the purpose of finding the equation for the location of the air car on the air track? Is there any specific experiment or scenario that this equation will be used for?

Assuming that the purpose is to determine the position of the front of the air car on the air track when the average velocity is L/t, here is my response:

1. The equation for the location of the air car on the air track when the average velocity is L/t is x = d + (L/t)t, where x is the location of the front of the car, d is the initial distance of the car from the photogate, and t is the time the car blocks the photogate.

2. Yes, the average speed and average velocity are not exactly the same. Average speed is the total distance traveled divided by the total time taken, whereas average velocity is the displacement divided by the total time taken. In this scenario, since the car is moving in a straight line, the average speed and average velocity will be the same.

3. Your first guess is correct. The equation for average velocity is v = (xf - xi)/t, where v is the average velocity, xf is the final position, xi is the initial position, and t is the time taken. In this case, xf = d + L and xi = d, so the equation becomes v = L/t. Since the average velocity is given as L/t, we can substitute it in the equation and get x = d + (L/t)t. This equation does not require the acceleration to be found.

In conclusion, the equation for the location of the air car on the air track when the average velocity is L/t is x = d + (L/t)t. This equation can be used to determine the position of the front of the car when the average velocity is known. It does not require the acceleration to be found. However, if the purpose is to study the motion of the car and its acceleration, then a different set of equations would be needed.