Find the total displacement on a velocity-time graph

In summary, the total displacement of the graph is 14m. This is calculated by taking the area under the curve and subtracting the negative displacement from the positive displacement. The first 2 seconds show the car moving backwards at a velocity of 4m/s, resulting in a negative displacement. The remaining time shows the car moving forwards at a velocity of 4m/s, resulting in a positive displacement. Thus, (-4m/s * 2s) + (4m/s * 4s) = -8m + 16m = 14m. This is because displacement is a vector quantity and is calculated by multiplying velocity and time.
  • #1
freefalling
5
0

Homework Statement


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
 

Attachments

  • physics.jpg
    physics.jpg
    22 KB · Views: 1,814
Physics news on Phys.org
  • #2
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.
 
  • #3
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.
 
  • #4
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?
 
  • #5
Yes, but I just don't get why it's 16-2...
 
  • #6
Take the difference between the area above the graph and the area below the graph.
 
Last edited:
  • #7
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?
 
  • #8
It doesn't say, the graph is all I got :( Thank you for your help!
 
  • #9
freefalling said:
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.
 
  • #10
So what you don't get is the reason one of the areas subtracts from the other?
 
  • #11
Yes :( like why it's 16-2 but not 16-6 (thats what makes sense to me)
 
  • #12
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?
 
  • #13
freefalling said:
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 =V×t
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.
 

1. What is a velocity-time graph?

A velocity-time graph is a graphical representation of an object's motion, showing the relationship between its velocity (speed and direction) and time. The slope of the graph represents the object's acceleration, while the area under the graph represents the object's displacement.

2. How do you find the total displacement on a velocity-time graph?

To find the total displacement on a velocity-time graph, you need to calculate the area under the graph. This can be done by dividing the graph into smaller shapes (triangles, rectangles, etc.) and calculating their individual areas. Then, add all the individual areas together to get the total displacement.

3. What does a positive or negative area on a velocity-time graph represent?

A positive area on a velocity-time graph represents a positive displacement, meaning the object is moving in the positive direction. A negative area represents a negative displacement, meaning the object is moving in the negative direction (opposite of positive).

4. How does the slope of a velocity-time graph relate to an object's acceleration?

The slope of a velocity-time graph represents the object's acceleration. A steeper slope indicates a higher acceleration, while a flatter slope indicates a lower acceleration. A horizontal line (zero slope) represents a constant velocity, and a vertical line (infinite slope) represents an instantaneous change in velocity (such as during free fall).

5. Can you determine an object's displacement at a specific time on a velocity-time graph?

Yes, you can determine an object's displacement at a specific time on a velocity-time graph by finding the area under the graph up to that specific time. This is known as the object's displacement at a specific time interval. However, to find the total displacement over a specific time period, you need to calculate the area under the entire graph.

Similar threads

Position from velocity time graph
    • Introductory Physics Homework Help
Replies
4
Views
2K
Free vibration in 2DOF spring mass systems
    • Introductory Physics Homework Help
Replies
2
Views
577
Needs clarity on this displacement vector diagram
    • Introductory Physics Homework Help
Replies
3
Views
408
Understanding a Velocity-Time Graph
    • Introductory Physics Homework Help
Replies
14
Views
3K
Finding Displacement based on Velocity Graph
    • Introductory Physics Homework Help
Replies
11
Views
2K
Kinematics: Find average speed from displacement-time graph.
    • Introductory Physics Homework Help
Replies
10
Views
4K
Finding acceleration from Velocity vs Position graph
    • Introductory Physics Homework Help
Replies
2
Views
2K
Virtual displacement for a block sliding down a wedge
    • Introductory Physics Homework Help
Replies
3
Views
1K
Distance-time graph of a ball throw vertically up from a fixed point
    • Introductory Physics Homework Help
Replies
5
Views
256
Solving Distance Traveled Using Formulas vs. Graphs
    • Introductory Physics Homework Help
Replies
6
Views
785

Hot Threads

Recent Insights

Back
Top