[Motion Problem] When is displacement equal?

  • Thread starter Ticklez_Panda
  • Start date
  • Tags
    Displacement
In summary, the objects have a constant acceleration, but line:B changes its velocity constantly whereas line:A changes only once it has reached its final velocity.
  • #1
Ticklez_Panda
6
0
The figure below shows a velocity versus time plot for two objects; they are at the same point at
t = 0.
3-p-005.gif


Homework Statement


Do the objects ever have the same displacement? If so, when does that happen?


Homework Equations


Δd = ((velocityi+ velocityf) / 2) Δt
acceleration = change in velocity / change in time
Δd(A) = 100Δt + 0.5 (accel. Δt^2)
Δd(B) = 30Δt + 0.5 (accel. Δt^2)

The Attempt at a Solution


Looking at the graph, I can tell that line:B has a constantly changing velocity and therefore a constant acceleration. Line:A is different, it resembles a quadratic and displays the characteristic of a non-constant (time dependent acceleration). I know that the displacement between the two lines will be the same when the sum of initial and final velocity are equal given by my displacement equation.

Could someone please help to guide me towards this solution? It seems incredibly simple but I just can't grasp it. :(
 
Last edited:
Physics news on Phys.org
  • #2
You need to use the more fundamental relationship between distance and speed: in this case,
d(t) = ∫v(t)dt. Use that for both curves.

To get a precise numerical answer you have to know what vA(t) is. Looks like a parabola to me, so derive the equation for the A curve. The B curve is obvioulsy a straight line, as you said.

Otherwise you can try to approximate the answer graphically. Not easily, though.
 
  • #3
Representing the lines in equations

So I know that the line:B can be graphed with y = mx + b, which I resolved to equal y = 0.5x + 30. For line:A, I am having trouble representing it in an equation form. It appears to be quadratic, therefore it should be represented with y = ax^2 + bx + c. Where c = 100, I equated a to equal 1.6 using the first 50 seconds as a basis. I can't seem to find the solution to the second equation but I can estimate the x-int to be about 75 and 145, with a y-int of 100. What am I doing wrong here?
 
  • #4
A parabola would be v-v0 = k(t - t0)^2. Try that.
v0 = -10 is pretty obvious but t0 less so. k somewhere around 0.01, t0 around 110 s.
 
  • #5


Your approach is correct. To find when the displacement is equal, you need to set the two displacement equations equal to each other and solve for Δt:

Δd(A) = Δd(B)
100Δt + 0.5 (accel. Δt^2) = 30Δt + 0.5 (accel. Δt^2)
70Δt = 0
Δt = 0

This means that the objects have the same displacement at t = 0, which is when they are both at the same point. This makes sense because the graph shows that both objects start at the same point at t = 0.
 

1. When is displacement equal to zero?

Displacement is equal to zero when an object returns to its original position, meaning there is no change in its position from the starting point.

2. How is displacement different from distance traveled?

Displacement refers to the change in an object's position from its starting point to its ending point, while distance traveled refers to the total length of the path an object has taken.

3. Can displacement be negative?

Yes, displacement can be negative if an object moves in the opposite direction from its starting point. It indicates that the object's ending position is behind its starting position.

4. Is displacement always equal to the shortest distance between two points?

No, displacement is not always equal to the shortest distance between two points. It only takes into account the change in position, not the actual distance traveled.

5. How is displacement calculated?

Displacement can be calculated by subtracting the initial position from the final position. It is represented by the symbol Δx, which stands for change in position.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
2K
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
5K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
2K
Back
Top