Find the total displacement on a velocity-time graph


by freefalling
Tags: displacement, graph, velocitytime
freefalling
freefalling is offline
#1
Jun16-12, 05:14 PM
P: 5
1. The problem statement, all variables and given/known data
What is the total displacement of the graph? (see the picture)
The answer is 14m. Can you please show me how to do it?
Thank you for your help! :D
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Martinet
Martinet is offline
#2
Jun16-12, 05:43 PM
P: 28
Give it a crack first. Write up your attempt here :) It's against the rules of the forum to give help without the asker proving they're not just being a slacker.
freefalling
freefalling is offline
#3
Jun16-12, 05:56 PM
P: 5
I calculated the area under the curve and then I subtracted 2 from 16 and I got 14. But the steps don't make sense to me.

Martinet
Martinet is offline
#4
Jun16-12, 06:09 PM
P: 28

Find the total displacement on a velocity-time graph


What exactly doesn't make sense?
Do you understand what the product of a velocity-time actually is graph is?
Are you aware of the vector nature of velocity and displacement?
freefalling
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#5
Jun16-12, 06:15 PM
P: 5
Yes, but I just don't get why it's 16-2...
Xisune
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#6
Jun16-12, 07:26 PM
P: 54
Take the difference between the area above the graph and the area below the graph.
azizlwl
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#7
Jun16-12, 07:41 PM
P: 961
Taking forward as positive.

At starting point, your reversing speed is 4m/s and you applied brake and the car slowly reducing its speed till 2 secs later your speed is zero.

At time=2 sec and your speed is zero, you start forward for 2 secs where you're back to your starting point.

And you continue moving forward at constant speed of 4m/s for 2 sec.

At t=6sec you applied brake and the car move for 2 sec and reach final speed of 2m/s.

What is your distance from the starting point?
freefalling
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#8
Jun16-12, 08:06 PM
P: 5
It doesn't say, the graph is all I got :( Thank you for your help!
azizlwl
azizlwl is offline
#9
Jun16-12, 09:16 PM
P: 961
Quote Quote by freefalling View Post
It doesn't say, the graph is all I got :( Thank you for your help!
That's what you should roughly say when looking and interpreting the graph. Hope it is helpful.
Martinet
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#10
Jun17-12, 12:50 AM
P: 28
So what you don't get is the reason one of the areas subtracts from the other?
freefalling
freefalling is offline
#11
Jun17-12, 01:10 AM
P: 5
Yes :( like why it's 16-2 but not 16-6 (thats what makes sense to me)
Martinet
Martinet is offline
#12
Jun17-12, 01:32 AM
P: 28
Okay. You understand that, as a vector quantity, displacement is a number value representing distance combined with an angle, measured from a certain point. This problem takes away the need for an angle, because the car can only move in two directions - forward and backward. This makes it easy. We call forward positive and backward negative.

So, if I move forward 1m from point X, my displacement is 1m.
If I move backward 1m from point X, my displacement is -1m.

Knowing this, and knowing that we have a velocity-time graph, we need to look at how velocity relates to displacement. Velocity is also a vector, and so - in this case - must either be positive or negative. You can think of it as speed with a direction. Again forward is positive and backward is negative. So. Given that velocity = distance / time:

If I move 1m forward and it takes me 1 second, my velocity is 1 metre per second (m/s)
If I move 1m backward and it takes me 1 second, my velocity is -1m/s.

If I move for 2 seconds at 1m/s, my displacement is 2m.
If I move for 2 seconds at -1m/s, my displacement is -2m.

Knowing all this, we can look at the graph. Clearly you know that the area under the graph gives displacement. If you don't know why:

We're multiplying the x-axis by the y-axis.
The x-axis is time and the y-axis is velocity.
Therefore, x = t, and y = d / t.
x*y = (d / t) * t = d.
d = displacement.

So looking at the graph we see that there are two distinct sections. One where velocity is negative (backwards) and one where velocity is positive (forwards). This translates into the car reversing from Point A and then moving forwards towards point A and continuing.

Clearly, because (v / t) * t = d, if we take a negative velocity and multiply it by time we'll get a negative displacement. Accordingly we will get a positive displacement from a positive velocity.

Do you understand why it is subtracted now?
azizlwl
azizlwl is offline
#13
Jun17-12, 02:11 AM
P: 961
Quote Quote by freefalling View Post
Yes :( like why it's 16-2 but not 16-6 (thats what makes sense to me)
Velocity is a vector and time is a scalar quatity,
Displacement =Vt
Thus displacement is a vector quantity.

First 2 sec. the velocity is negative. Thus displacement is negative we call it d1
The rest is positive velocity. Positive displacement.d2.

Total displacement=(-)d1+d2.


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