Finding Displacement based on Velocity Graph

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  • #1
Lori

Homework Statement


A goalie moves in a straight line and her velocity is described by the graph shown. How far away is the goalie at the time t=2.0 from where she was at t =0?

The description of the velocity/time graph:
initial velocity is 3.0 m/s and increases to 4.0 m/s in 1.2 seconds (slope m =1/(1.2)). From 1.2 seconds to 2 seconds, the velocity decreases to 0 (the slope seems to be m= -2)

Homework Equations


Displacement = area of undercurve.

The Attempt at a Solution


I was given the graph, so i thought displacement would be to find the area under the lines...but apparently not.
I subtracted the area that forms the square and the triangle to find the displacement and got 2.6 m/s



Can someone please tell me what I did wrong?! :(
 

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  • #2
Ray Vickson
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Homework Statement


here's the pic of the graph:

https://imgur.com/a/rq1Xb

Homework Equations


Displacement = area of undercurve.

The Attempt at a Solution


[/B]
I subtracted the area that forms the square and the triangle to find the displacement, but it's the wrong answer.

My work is shown in the picture...

Can someone please tell me what I did wrong?! :(
Vert few, if any, helpers will look at your attachment until you take the trouble to post it with correct orientation. Lying sideways on my desk in order to read it is not an option.

Anyway, the graph is simple enough to describe in words and just type out here.
 
  • #3
cnh1995
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@Lori, why are you subtracting A2 from A1+A3?
 
  • #4
Lori
@Lori, why are you subtracting A2 from A1+A3?
I thought since the goalie is going in the negative direction that i should subtract them... Am i suppose to just add them all up?
 
  • #5
cnh1995
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Am i suppose to just add them all up?
Yes.
 
  • #6
I like Serena
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I thought since the goalie is going in the negative direction that i should subtract them... Am i suppose to just add them all up?
Suppose the goalie were going in the negative direction, what would that mean for his velocity?
Wouldn't his velocity be negative as well?
 
  • #7
Lori
Can someone explain how the displacement can be found from calculating the area under the graph? I don't understand how
 
  • #8
cnh1995
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Can someone explain how the displacement can be found from calculating the area under the graph? I don't understand how
Have you studied calculus?
 
  • #9
Ray Vickson
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Homework Statement


A goalie moves in a straight line and her velocity is described by the graph shown. How far away is the goalie at the time t=2.0 from where she was at t =0?

The description of the velocity/time graph:
initial velocity is 3.0 m/s and increases to 4.0 m/s in 1.2 seconds (slope m =1/(1.2)). From 1.2 seconds to 2 seconds, the velocity decreases to 0 (the slope seems to be m= -2)

Homework Equations


Displacement = area of undercurve.

The Attempt at a Solution


I was given the graph, so i thought displacement would be to find the area under the lines...but apparently not.
I subtracted the area that forms the square and the triangle to find the displacement and got 2.6 m/s



Can someone please tell me what I did wrong?! :(
You would subtract areas if her velocity switched from + to - , but that is not what is happening here. Here velocity remains > 0, but decreases between t = 1.2 and t = 2 seconds. The acceleration switches from + to - , but that is a different story.
 
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  • #10
I like Serena
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Can someone explain how the displacement can be found from calculating the area under the graph? I don't understand how
Suppose we drive a car at constant speed.
Then the distance covered is the speed times the time that we drive, isn't it?
Doesn't that correspond to the area of the rectangle that is under the speed-time graph?
Now suppose we slow down a bit and keep driving for the same time.
Then the total distance increases by the area of the second rectangle, doesn't it?

More generally this is also true if our speed changes continuously.
The distance covered is the area under the speed-time graph.

To reduce the total distance, we need to put the car in reverse and drive backwards -- that is, we need negative speed.
 
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  • #11
Lori
Suppose we drive a car at constant speed.
Then the distance covered is the speed times the time that we drive, isn't it?
Doesn't that correspond to the area of the rectangle that is under the speed-time graph?
Now suppose we slow down a bit and keep driving for the same time.
Then the total distance increases by the area of the second rectangle, doesn't it?

More generally this is also true if our speed changes continuously.
The distance covered is the area under the velocity-time graph.

To reduce the total distance, we need to turn the car around and drive back -- that is, we need negative velocity.
I always forget that positive velocity even if it's decreasing that it is still moving in the positive direction if it's above the x axis. Thanks for that!!
 
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  • #12
cnh1995
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Have you studied calculus?
Mathematically, velocity v(t) is the rate of change of displacement with time i.e.
v(t)=dx/dt

∴∫v(t) dt=∫dx

i.e. x(t)=∫v dt

Since integration of v w.r. time represents area under the v-t curve, displacement is the area under v-t curve (and its slope gives the instantaneous acceleration dv/dt).

Similarly, area under acceleration-time curve would give you velocity.
 

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