# A or B? Increase in Velocity Backwards & Acceleration Forward

• paulimerci
in summary, the object moves from zero to 1 second with a velocity in the negative direction, slows down, and comes to a stop.
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?

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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.

paulimerci
paulimerci said:
No equations!
You don't know any equations relating velocity, time and position under (piecewise) constant acceleration?

haruspex said:
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.

paulimerci said:
? 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
haruspex said:
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?

paulimerci said:
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 !

paulimerci said:
So what is really happening at t=1sec?
What you already stated:
paulimerci said:
comes to rest at t = 1
paulimerci said:
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?

phinds said:
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.

paulimerci said:
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
haruspex said:
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!

phinds said:
Well, the answer you have been give is VERY clear. Read it carefully and pay attention.
Thank you!

andrewkirk said:
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!

paulimerci said:
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.

PeroK said:
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.

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?

paulimerci said:
Is it possible to find t without integral?
You don't need integration to find the area of a triangle or rectangle.

PeroK said:
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?

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

paulimerci
PeroK said:
Yes, that's right. You can see visually that area
II is larger than area I (twice as large in fact).
Thank you!

## What does "increase in velocity backwards" mean?

An increase in velocity backwards means that an object is moving in the reverse direction and its speed is increasing. Essentially, the object is accelerating in the negative direction relative to a chosen frame of reference.

## How is acceleration forward different from an increase in velocity backwards?

Acceleration forward refers to an increase in the speed of an object in the positive direction of the chosen frame of reference. In contrast, an increase in velocity backwards means the object is speeding up in the negative direction. Both involve changes in speed, but in opposite directions.

## Can an object have both an increase in velocity backwards and acceleration forward simultaneously?

No, an object cannot have both an increase in velocity backwards and acceleration forward at the same time. These two conditions are mutually exclusive as they involve motion in opposite directions.

## How do you calculate acceleration if you know the increase in velocity backwards?

Acceleration is calculated as the change in velocity divided by the time over which the change occurs. If you know the increase in velocity backwards, you would use the negative value of the velocity change in your calculation. For example, if the velocity backwards increases by 5 m/s over 2 seconds, the acceleration would be -5 m/s divided by 2 s, resulting in -2.5 m/s².

## What are some real-world examples of "increase in velocity backwards" and "acceleration forward"?

A real-world example of an increase in velocity backwards is a car reversing and speeding up. An example of acceleration forward is a car moving forward and increasing its speed. Both scenarios involve changes in speed but in opposite directions.

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