Uniform Motion 2?

Question

for uniform motion, the line in a d-t graph is horizontal and in a v-t graph, the line is horizontal. I did an experiment where i have to push a cart and calculate the position and velocity of it for 1.0 second. However ,when I graphed the data, the line was not straight. What are some unavoidable sources of experimental error that might have cause my graph to differ from what was expected?

Answer 1

If I read your question correctly, you should have zero velocity; so your v-t graph would show a horizontal line stemming from the t = 0 origin, but coincident with the T axis. Why?

Because you said "the line in a d-t graph is horizontal". That means d is a constant, say D, no matter how much time elapses along your T axis. Thus, as v = deld/delt = 0/delt = 0; so the velocity of the cart v = 0 for any time elapsed. [NB: del* means change in *, whatever * is. In this case * is distance d and time t. A change in distance over a change in time lapsed is called velocity.]

I think your problem is that d <> constant, in fact deld > 0 Therefore, as time t increases, d should also be increasing. That is, the line on the d-t graph should be sloping upward to the right as you move along the T axis. Why?

Because your cart started at d = 0 at t = 0 and ended up at d = D some time (T) later. d < D in this case. Thus, D - d > 0 and distance increased with time (which is why the slope is upward and to the right).

In which case v = deld/delt = (D - d)/(T - t) > 0; where T is the end time when your cart is at D and t = 0 when you started timing the move of the cart. Now, here's the cool part, a straight line, like for the d-t graph, can be written in the form y = mx + b; where m is the slope of the line and b is the Y intercept when x = 0.

Thus, we can write the equation for the line on the d-t graph as d = (deld/delt)t + b; where deld/delt is the slope (m) of the line. And, hey, as deld/delt = v, we can write d = vt + b. And if d = 0 when t = 0, then b = 0 as well. This leaves us with d = vt TA DA, distance equals velocity times time. You just derived a SUVAT equation written in your textbook.