Displacement & distance on velocity vs time graph

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  • #1
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332756033.jpg


How do I calculate final displacement and the total distance traveled in the graph above.

I know displacement is independent of the pathway so the distance traveled on the last point is 44 m and the first point is 0 m.

I also know for distance that you have to calculate the area of the triangle above the x-axis (1/2*b*h) but I'm not sure how to calculate the area under the x-axis.
 

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  • #2
LowlyPion
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How do I calculate final displacement and the total distance traveled in the graph above.

I know displacement is independent of the pathway so the distance traveled on the last point is 44 m and the first point is 0 m.

I also know for distance that you have to calculate the area of the triangle above the x-axis (1/2*b*h) but I'm not sure how to calculate the area under the x-axis.

To be simple minded about it you can just say 1/2 base times height above and sub out 1/2 base times height of the areas below the x-axis as well as the base times height of the rectangular area.

For instance above x-axis positive area is 1/2 (5 - 1) * (8).
Below the line be careful about which triangles you choose and which rectangles to fully account for all the area.

The important thing is you understand what the area of this function represents. It is the integral of F(t) which is Velocity(t) from 0 to 11 seconds. The integral of velocity is ... distance.

If you needed acceleration, that would be the derivative of this Velocity(t), and the slopes of those lines would be the acceleration during those time intervals.
 
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  • #3
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For area under the x-axis I got 34 so 34 + 16 = 50 m?

This is what I did: 12 x (8-5) x (1/2) = 18

area of rectangle: 1 x 12 = 12 - [(1/2) * 1 * 8] = 8

4 x (11 - 9) = 8

18 + 8 + 8 = 34
 
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  • #4
LowlyPion
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For area under the x-axis I got 34 so 34 + 16 = 50 m?

This is what I did: 12 x (8-5) x (1/2) = 18

area of rectangle: 1 x 12 = 12 - [(1/2) * 1 * 8] = 8

4 x (11 - 9) = 8

18 + 8 + 8 = 34

That's correct.
 
  • #5
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Okay let me try again.

Is it 26 m?

(26 + 16 = 42)?
 
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  • #6
LowlyPion
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Okay let me try again.

Is it 26 m?

(26 + 16 = 42)?

I think 50 is the right answer. I originally posted I thought it was wrong by 1 and then realized the y scale was 2 per unit and changed my post. Maybe you were quick enough to catch my first posting of it before I corrected (edited) myself. If so, sorry for any inconvenience.
 
  • #7
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No problem.

So for displacement would you use [v1 + v2 / 2] * t = d?

[0 + -44]/2 x [11-1] =

-22/10 = - 2.2 m?

This does not seem right.
 
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  • #8
LowlyPion
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No problem.

So for displacement would you use [v1 + v2 / 2] * t = d?

[0 + -44]/2 x [11-1] =

-22/10 = - 2.2 m?

This does not seem right.

Wait a minute. A big mistake. We made it together.

The positive area above V=0 is 16. The area below V=0 is 34 but that is negative. That means there is negative total displacement of 16 - 34 = -18.
The distance traveled is 50 m.

Now the 44 m you are talking about I am not understanding where that comes from.
 
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  • #9
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Wait a minute. A big mistake. We made it together.

The positive area above V=0 is 16. The area below V=0 is 34 but that is negative. That means there is negative total displacement of 16 - 34 = -18.

Now the 44 m you are talking about I am not understanding where that comes from.

I got -44 by multiplying -4.0 m/s * 11.0 s = - 44 m.
 
  • #10
LowlyPion
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I got -44 by multiplying -4.0 m/s * 11.0 s = - 44 m.

But -4 m/s is only the velocity at 11 seconds. The graph is the locus of all previous velocities at all previous points back to t = 0.
 
  • #11
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But -4 m/s is only the velocity at 11 seconds. The graph is the locus of all previous velocities at all previous points back to t = 0.


Got it. Thanks.
 

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