Help with experiment...
Right now I'm doing a lab report on the motion of a cart moving down an incline. So far, I've got my purpose, materials, and procedure written down. I'm currently doing the response questions. There are 10 of them, and I've got 6 so far. Two of them I'll be fine with, but one of them in particular right now is puzzling me.
The question I need help on says: Sketch the shapes of the position-time, velocity-time, and acceleration time graphs for the cart rolling down the ramp.
A previous procedure says: Plot a graph of velocity against time for the cart by plotting the average velocities at half-time intervals (0.05s, 0.15s, etc).
I graphed my results and got the following (see attached). My question is, am I supposed to use half-time intervals or normal-time intervals for the three graphs? What’s the purpose of half time? It just doesn’t make sense to ask for the same graph twice.
How are you measuring speed, is it the distance moved in a fixed time interval? If so, that speed may be regarded as the speed at the mid point of the interval (which is what I think the question is trying to say).
Originally Posted by shadow6
The velocity was calculated by dividing the displacement during a fixed time interval buy 0.1 (the time). The attached is a chart containing my calculations. The only thing that's confusing me about this experiment is the half time aspect of it.
Another question asks, "For the same time interval, how did the area beneath the velocity-time graph compare with the displacement of the cart? How should it compare?" My answer was as follows:
(.5vt) = (.5vt(15m/s)(0.05s)) = 0.375m.
The displacement of the cart calculated by finding the area beneath the velocity-time graph is 0.375cm [down]. The area beneath the graph (distance) over the 0.05s interval did not equal the distance in the graph (0.375cm does not equal 1.5cm). To obtain the average velocity for the graph, the displacement (1.5cm) was divided by 0.1s. This gave an average velocity of 15cm/s. In order to get the displacement from the velocity, the operation must be done in reverse; (15cm/s)(0.1s) = 1.5cm. This is not what is being done with the graph. On the graph, half time units are being used as opposed to full time. So already, it can be determined that the velocity will at least be half of 1.5cm; that being 0.75cm. In addition, this answer must be multiplied by 0.5 (one half) due to the area being a triangle, meaning that the displacement will be 0.375cm. Therefore, the displacement determined from the Velocity-Half Time graph will always be one fourth of the displacement given on the chart. My hypothesis predicting that they would be equivalent was wrong.
This answer doesn't seem right though. I mean, based on the calculations, the answer seems right. But shouldn't the displacement under the graph always equal the displacement on the chart?