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Old Mar 29th 2011, 11:25 AM   #1
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Area under the velocity time graph

This assingment is due tomorrow
I found the area under a irregular shape velocity-time graph.

I counted 37 full squares and my area for each square is 1.5 m/s*3s. Hence my displacement is 166.5 m BUT it is noted that the area under the graph is the displacement. I am asked to find the displacement what i wanted to know is if i have to use a new formula to find distance or is this method fine for calculating the distance?

I plotted the graph on excel 2007 and it wouldnt load to get a general idea let me just give you guys the points

Time 0 4 6 8 12 15 18 20 22 23 26 30
Velocity m/s 0 2.5 4.0 6.5 8.0 8.0 8.5 9.0 7.5 5.2 4.0 2.5

Last edited by purplec16; Mar 29th 2011 at 12:05 PM.
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Old Mar 29th 2011, 04:40 PM   #2
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Originally Posted by purplec16 View Post
This assingment is due tomorrow
I found the area under a irregular shape velocity-time graph.

I counted 37 full squares and my area for each square is 1.5 m/s*3s. Hence my displacement is 166.5 m BUT it is noted that the area under the graph is the displacement. I am asked to find the displacement what i wanted to know is if i have to use a new formula to find distance or is this method fine for calculating the distance?

I plotted the graph on excel 2007 and it wouldnt load to get a general idea let me just give you guys the points

Time 0 4 6 8 12 15 18 20 22 23 26 30
Velocity m/s 0 2.5 4.0 6.5 8.0 8.0 8.5 9.0 7.5 5.2 4.0 2.5
Yes, the integral of a velocity vs. time graph will be the displacement. To get distance you need to integrate over the speed vs. time graph. The difference is that to get the distance you need to integrate |v| over time, where |v| is the absolute value of the velocity v.

In your case they will be the same since your graph of v does not go below the time axis.

-Dan
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Old Mar 29th 2011, 07:05 PM   #3
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Alrighty... thank you so much...
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