Position from velocity time graph

In summary: The total distance it moved is 2 m, but it is not 2 m to the right, it is 2 m to the right minus the 2 m to the left it moved to get back to its starting point.)In summary, the particle's position at t = 2 seconds is 14 meters. The initial position is 14 meters and the velocity graph shows the particle moving with a velocity of 4t, where t is time in seconds. The displacement of the particle is equal to the area under the velocity graph, and the negative portion of the graph indicates movement in the opposite direction, resulting in a displacement of 0 meters. Using calculus, the displacement can be found by integrating the velocity function and adding it
  • #1
sp3sp2sp
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Homework Statement


A particle starts from x_0 = 14m at t_0 = 0 and moves with the velocity graph shown
What is the object's position at t = 2 s?

Homework Equations


Area under velocity time graph is the displacement of the particle.

The Attempt at a Solution


The part that confuses me is where the velocity goes negative. I think that I should assign negative value to the portion where v_x is negative.
So from t=1 to t=2: (1/2)(1*4) = 2m
from t=0 to t=1 it is negative: - (1/2)(1*4) = -2m
-2m+2m = 0m = no displacement?
Is this correct before I try using calculus to solve same problem? Thanks!

sorry I forgot the initial velocity = 14m/s.
So objects position after 2s is 0m + 14m = 14m

 

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  • #2
Correct, with one minor correction: Second to last line: Object's initial position is ## x_o=14 ## m. The initial velocity is not 14 m/sec.
 
  • #3
Thanks for the help. So for the calculus version I am again confused about what to do with the negative part.
First the equation for the velocity is v_x = 4t (because slope is 4/1 = 4)
So if I intergrate that I get x = 2t^2 + c
From x=1 to x= 2 is 1s: x = 2(1)^2 = 2m (Im not sure what the c represents)
I know that from x=0 to x=1 also = 1s, and that it is the same equation, but how do I represent the negative so that it will cancel, as like the algebra version? thanks
 
  • #4
These are definite integrals. See (3 lines) below. ## \\ ## The correct expression for ## v_x ## is ## v_x(t)=-4+4t ##.## \\ ## (You can write ## \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} ##, where you use any two points ## (x_1,y_1) ## and ## (x_2,y_2) ##. In this case ## y ## is ## v ##, and ## x ## is ## t ##). ## \\ ## The distance ## \Delta x ## it goes is ## \Delta x=\int\limits_{0}^{2} v_x(t) \, dt ##, (as previously mentioned, it's a definite integral, i.e. one that has its limits posted in the expression, and thereby there is no constant of integration to add to the result like there is with an indefinite integral). ## \\ ## Finally, ## x_{final}=x_o +\Delta x ##. ## \\ ## I'll let you perform the integral. ## \\ ## You could also write the integral as ## \Delta x=\int\limits_{0}^{1} v_x(t) \, dt +\int\limits_{1}^{2} v_x(t) \, dt ## if you want to sum these two parts separately.
 
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  • #5
sp3sp2sp said:
Thanks for the help. So for the calculus version I am again confused about what to do with the negative part.
First the equation for the velocity is v_x = 4t (because slope is 4/1 = 4)
So if I intergrate that I get x = 2t^2 + c
From x=1 to x= 2 is 1s: x = 2(1)^2 = 2m (Im not sure what the c represents)
I know that from x=0 to x=1 also = 1s, and that it is the same equation, but how do I represent the negative so that it will cancel, as like the algebra version? thanks

If positive velocities/displacements are to the right and negative ones to the left, your velocity graph has the particle moving to the left from time 0 to 1. The distance it moves left in that time is the magnitude of the area between the v-graph and the t-axis. (Note that the distance moved is positive and the magnitude of the area is also positive---however, the displacement is negative!) At t=1 the particle starts moving to the right, and between t=1 and t=2 the distance it moves right equals the area under the v-graph and the t-axis. So, between t = 1 and t = 2 the particle just moves back to its starting position.
 

1. How do you determine the position from a velocity time graph?

To determine the position from a velocity time graph, you need to find the area under the velocity curve. This area represents the displacement, or change in position, of the object over the given time interval.

2. What does the slope of a velocity time graph represent?

The slope of a velocity time graph represents the acceleration of the object. A steeper slope indicates a higher acceleration, while a flat slope represents constant velocity.

3. Can you have a negative position on a velocity time graph?

Yes, a negative position on a velocity time graph simply means that the object has moved in the opposite direction of the chosen reference point. It does not necessarily indicate a decrease in position.

4. How can you tell if an object is at rest or moving at a constant velocity on a velocity time graph?

If the velocity time graph is a straight horizontal line, the object is at rest. If the graph is a straight diagonal line, the object is moving at a constant velocity. A curved line indicates a changing velocity, and therefore, the object is not at rest.

5. How does the shape of a velocity time graph relate to the motion of an object?

The shape of a velocity time graph represents the motion of an object. A flat line represents constant velocity, while a curved line indicates a changing velocity. A steep slope indicates a higher acceleration, while a shallow slope represents a lower acceleration. A horizontal line at the x-axis represents a stationary object.

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