# Using a velocity-time graph to calculate accelerations, positions, etc.

## Homework Statement

using the diagram found at this link:

http:/www.physics.uoguelph.ca/phyjlh/1020/Asst_1/asst1_q1v.jpg"

a) Plot a graph which shows that the acceleration is constant within the errors of measurement.
b) What is the acceleration of the car?
c) What was the velocity of the car at t = 0?
d) If the acceleration remains constant over the next 5 seconds what will be the velocity of the car at t = 9s?
e) What will be the position of the car at the time given in Part d?

v=d/t , a = v/t

## The Attempt at a Solution

so far I've really only tried part a) and I just can't seem to find the answer or any relevant help. What I have done is I have used the 25 m calibration to derive a 1 m = 0.372 cm scale for the graph. after doing ths I have measure all the distances at every 1 s interval. The measurements I have: 0.91 cm = 0.335 m (t=1 s) ; 2.75 cm = 1.02 m (t=2) ; 4.69 cm = 1.74 m (t=3) ; 6.45 cm = 2.40 m (t=4).

I have found velocities for all of these values: t 1 = 0.335 m/s ; t 2 = 0.510 m/s ; t 3 = 0.580 m/s ; t 4 = 0.600 m/s

First of all, are these values even correct? Because when I plot them I get nothing close to a linear graph. Are my errors of measurement that off? Ultimately, I'm just really confused and need some help starting this problem correctly. I feel that once I have a grasp on how to go about doing part a) correctly everything would fall into place. This question is worth 2% of my final mark so I'm kind of freaking out.

Any help is appreciated.

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Math Is Hard
Staff Emeritus
Gold Member

Sorry about that. Could you maybe just copy and paste into your browser?

www.physics.uoguelph.ca/phyjlh/1020/Asst_1/asst1_q1v.jpg[/URL]

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Here's the picture diagram attached

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Delphi51
Homework Helper
The link doesn't work because part of it is missing (replaced by ...).

It appears that you have distance and time points and want a graph to find acceleration. If so, you must take a look at your accelerated motion formulas that have distance and time in them. And no velocity. If the initial velocity is zero, you could use d = .5at^2.
The idea is to compare it to y = mx + b and see if you can come up with a way to make your data form a straight line with a slope that is related to the acceleration. Forget the b - not needed - and compare
d = .5at^2 with y = mx
Clearly you need to put d on the y-axis and t^2 on the x-axis. That leaves b corresponding to .5a. So if that graph turns out to be a straight line, you have accelerated motion and the slope is .5a.

Of course the data will not form a perfect straight line because of experimental error. You have to judge if it looks more like a straight line than any other pattern. And the wise experimenter never says "it is a straight line" but rather try to estimate error bars for the measurements and see if it "is a straight line to within experimental error". Finally, do a line of best fit and get its slope - and a "steepest" line that still fits the data to within the error bars. Report your best line slope plus or minus the difference between it and the "steepest" line. Perhaps you have more sophisticated ways to do this with a calculator or computer that finds the line of best fit and its standard deviation.