At what time after t=0 does the object again pass its initial position?

[paulimerci]

Homework Statement:

The graph above shows the velocity versus time for an object moving in a straight line. At what time after t = 0 does the object again pass through its initial position?
(A) 1 s (B) Between 1 and 2 s (C) 2 s (D) Between 2 and 3 s

Relevant Equations:

No equations!

I've understood that between time t=0 to t=1 sec (moving backward), the object is moving with increasing velocity in the negative direction, slows down, and comes to rest at t = 1 sec. At t = 1 sec, the object returns to its starting position, briefly rests, and then begins to accelerate (moving forward). My answer to this question is A, and the worksheet answer says B. Which is correct?

 

[andrewkirk]

The worksheet is correct.
The graph shows velocity, not position, so at t=1, all of the object's velocity to date has been in the same direction, the negative direction. The object can't reach the starting position until its velocity has been in the positive direction long enough to backtrack over all the negative-direction ground it covered while it had negative velocity.
So what you need to find is the value of t such that the integral of the curve from 0 to t is zero.

 

[haruspex]

No equations!

You don't know any equations relating velocity, time and position under (piecewise) constant acceleration?

 

[paulimerci]

You don't know any equations relating velocity, time and position under (piecewise) constant acceleration?

I know, but do those equations help to solve the question above? Why can't the answer be t = 1 second? because that's where the object changes its direction.

 

[haruspex]

? Why can't the answer be t = 1 second?

For the reason @andrewkirk gave. For the first second all the movement has been in the negative direction. How can it be back at the start point?

 

[paulimerci]

For the reason @andrewkirk gave. For the first second all the movement has been in the negative direction. How can it be back at the start point?

So what is really happening at t=1sec? Was my interpretation above is wrong?

 

[phinds]

So what is really happening at t=1sec? Was my interpretation above is wrong?

Do you think the answer is going to change just because you keep asking the same question? READ POST #2 !

 

[haruspex]

So what is really happening at t=1sec?

What you already stated:

comes to rest at t = 1

that's where the object changes its direction

Walk to your front door and back. How long before you came to rest momentarily and then changed direction? How long before you returned to where you started?

 

[paulimerci]

Do you think the answer is going to change just because you keep asking the same question? READ POST #2 !

I understand the answer won’t change, but I want to understand it more clearly.

 

[phinds]

I understand the answer won’t change, but I want to understand it more clearly.

Well, the answer you have been give is VERY clear. Read it carefully and pay attention.

 

[paulimerci]

What you already stated:


Walk to your front door and back. How long before you came to rest momentarily and then changed direction? How long before you returned to where you started?

It's just a simple concept. I made things more difficult for myself. I think I understood. Thank you!

 

[paulimerci]

Well, the answer you have been give is VERY clear. Read it carefully and pay attention.

Thank you!

 

[paulimerci]

The worksheet is correct.
The graph shows velocity, not position, so at t=1, all of the object's velocity to date has been in the same direction, the negative direction. The object can't reach the starting position until its velocity has been in the positive direction long enough to backtrack over all the negative-direction ground it covered while it had negative velocity.
So what you need to find is the value of t such that the integral of the curve from 0 to t is zero.

Thank you!

 

[PeroK]

Homework Statement:: The graph above shows the velocity versus time for an object moving in a straight line. At what time after t = 0 does the object again pass through its initial position?
(A) 1 s (B) Between 1 and 2 s (C) 2 s (D) Between 2 and 3 s
Relevant Equations:: No equations!

I've understood that between time t=0 to t=1 sec (moving backward), the object is moving with increasing velocity in the negative direction, slows down, and comes to rest at t = 1 sec. At t = 1 sec, the object returns to its starting position, briefly rests, and then begins to accelerate (moving forward). My answer to this question is A, and the worksheet answer says B. Which is correct?

The displacement (signed area under a velocity time graph) is clearly non-zero between $0$ and $1$ seconds.

You should have been looking for a time when the signed area under the graph is zero.

 

[paulimerci]

The displacement (signed area under a velocity time graph) is clearly non-zero between $0$ and $1$ seconds.

You should have been looking for a time when the signed area under the graph is zero.

Correct! I should have done that. Thank you. I messed up with the velocity vs. time and position vs. time graphs.

 

[paulimerci]

I’ll take back what I said in post #1. I’m sorry. I understood a way more better now. I tried to plot a dual position vs time graph and it made lot of sense.
Is it possible to find t without integral?

 

[PeroK]

Is it possible to find t without integral?

You don't need integration to find the area of a triangle or rectangle.

 

[paulimerci]

You don't need integration to find the area of a triangle or rectangle.

Okay, the area under the velocity vs. time graph gives displacement.

Area I ( Area of two triangles between t=0s and t=1s)
Displacement = -0.5m
Area II (Area of triangle between t=1s and t=2s)
Displacement =1m
Area III ( Area of rectangle between t=2s to t=3s)
Displacement = 2m
Area IV (Area of triangle between t=3s to t=4s)
Displacement = 1m
Conceptually, I understand. The area under the curves I and II is where the object moves from negative displacement to positive displacement (-0.5, -0.4, -0.3, -0.2, -0.1, 0,1), so it shows after t = 0 that the object crossed its initial position between 1 and 2 s. Right?

 

[PeroK]

Yes, that's right. You can see visually that area
II is larger than area I (twice as large in fact).

 

[paulimerci]

Yes, that's right. You can see visually that area
II is larger than area I (twice as large in fact).

Thank you!