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

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In summary, the car had a velocity of 0 at time 0 and had an acceleration of .5a over the next 5 seconds. The car will have a velocity of 2.40 m at time 9s.
  • #1
ikasok
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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?


Homework Equations



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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  • #2
There's a problem with your link.
 
  • #3
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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  • #4
Here's the picture diagram attached
 

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  • #5
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.
 

1. What is a velocity-time graph?

A velocity-time graph is a visual representation of an object's velocity over time. It plots the velocity of an object on the y-axis and time on the x-axis. The slope of the graph at any given point represents the object's acceleration.

2. How can I calculate acceleration using a velocity-time graph?

To calculate acceleration from a velocity-time graph, you can use the formula a = (vf - vi) / t, where vf is the final velocity, vi is the initial velocity, and t is the time interval. Simply find the values on the graph and plug them into the formula.

3. Can I calculate the position of an object using a velocity-time graph?

Yes, you can calculate the position of an object using a velocity-time graph by finding the area under the graph. The area under the graph represents the displacement of the object. You can use the formula d = (vi + vf) / 2 * t to calculate the displacement.

4. What does a horizontal line on a velocity-time graph mean?

A horizontal line on a velocity-time graph means that the object's velocity is constant or there is no acceleration. This could mean that the object is at rest or moving at a constant speed.

5. Can I use a velocity-time graph for non-uniform motion?

Yes, a velocity-time graph can be used for non-uniform motion. In this case, the slope of the graph will not be constant, and the acceleration may change over time. You can still use the same formulas to calculate acceleration and position, but you may need to break the graph into smaller sections to get more accurate values.

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