# Positive or negative final velocity? Constant acceleration

• AnnaRiddle
So, to make my response make more sense, please read Chestermiller's response above as a response that I wrote in ignorance of Chestermiller's response.
AnnaRiddle

## Homework Statement

A car is moving with a constant velocity of 18 m/s for 5 seconds, if in the next 5 seconds it travels a distance of 40 m, what is its final velocity?

## Homework Equations

Δx= vit+ 1/2 at^2
vf= vi+at
vf^2 = vi^2 + 2a (Δx)

## The Attempt at a Solution

So I tried doing it different ways and I got -2 m/s in all of them, but by doing a velocity vs time graph it seems that +2 m/s is the right answer. Here's how I did it:

I first solved for acceleration from this formula:
Δx= vit+ 1/2 at^2, using 40 m as my Δx, 18 m/s as my starting velocity and 5 seconds for time, which gave me an acceleration of -4 m/s^2.
With that, I then plugged in vf= vi+at and got -2 m/s. But when I do it using this formula vf^2 = vi^2 + 2a (Δx) I get + 2 m/s.

My question is, is there a way to know exactly which direction it will go by only doing it algebraically? (I mean without doing the graph) also, why am I getting different signs? I can't seem to find the gap in my two processes.

Assuming a constant acceleration, I would use:

##d=\overline{v}t=\dfrac{v_i+v_f}{2}t##

Solving for ##v_f##, we obtain:

##v_f=\dfrac{2d}{t}-v_i##

Plug in the given values:

##v_f=\left(\dfrac{2\cdot40}{5}-18\right)\,\dfrac{\text{m}}{\text{s}}=-2\,\dfrac{\text{m}}{\text{s}}##

When you used the formula involving the squares of the initial and final velocities, you need additional information to determine whether to use the negative or positive root when solving for a squared value.

AnnaRiddle said:
But when I do it using this formula vf^2 = vi^2 + 2a (Δx) I get + 2 m/s.
Be careful here. What you get is ##v_f^2 = 4 \mathrm{m}^2/\mathrm{s}^2##, which means that ##v_f = \pm 2 \mathrm{m/s}##.

AnnaRiddle said:
by doing a velocity vs time graph it seems that +2 m/s is the right answer.
Can you post that graph?

AnnaRiddle said:
But when I do it using this formula vf^2 = vi^2 + 2a (Δx) I get + 2 m/s.
Square roots of 4 are both +2 and -2.
You need to use this equation
AnnaRiddle said:
vf= vi+at
where there is no square term. This gives Vf=-2m/s. Also, you can verify this by calculating the time at which the velocity becomes zero.

DrClaude said:
Be careful here. What you get is ##v_f^2 = 4 \mathrm{m}^2/\mathrm{s}^2##, which means that ##v_f = \pm 2 \mathrm{m/s}##.Can you post that graph?

Oh! So if I'm calculating both magnitude and direction of vf I have to use this vf= vi+at, but this vf^2 = vi^2 + 2a (Δx) would only give me the speed, I won't know for sure the direction with the data given.

I'll post the graph later because I'm a bit busy.

How far would it have traveled if its speed hadn't changed during those 5 seconds?

Chestermiller said:
How far would it have traveled if its speed hadn't changed during those 5 seconds?

90 m. How is that relevant? I know it traveled 90 m in the first part of the problem but it has a constant acceleration during the second portion.

AnnaRiddle said:
90 m. How is that relevant? I know it traveled 90 m in the first part of the problem but it has a constant acceleration during the second portion.
The actual distance was only 40 m. Do you think that means that it speeded up or slowed down?

AnnaRiddle said:
90 m. How is that relevant? I know it traveled 90 m in the first part of the problem but it has a constant acceleration during the second portion.

It is relevant because you need to figure out if it is "speeding up" or "slowing down" in the next 5 second period. That tells you if acceleration is positive or negative.

Note added in edit: I saw the new entry of Chestermiller only after I pressed the "enter" key to submit my posting. I often find that PF responses are delayed considerably before they appear on my screen, and that often leaves me looking like one of those helpers who ignore other responses.

Last edited:

## 1. What is the difference between positive and negative final velocity?

Positive final velocity refers to a direction of motion that is towards the positive direction of a coordinate system, while negative final velocity refers to a direction of motion that is towards the negative direction of a coordinate system.

## 2. How is final velocity calculated?

Final velocity can be calculated by adding the initial velocity to the product of acceleration and time. This can be represented by the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

## 3. What does constant acceleration mean?

Constant acceleration means that the velocity of an object is changing at a constant rate. This can be represented by a straight line on a velocity-time graph, where the slope of the line represents the acceleration.

## 4. How is constant acceleration related to final velocity?

Constant acceleration is directly related to final velocity, as the final velocity of an object is determined by its initial velocity, acceleration, and time. The acceleration remains constant throughout the motion, resulting in a change in velocity that can be calculated using the equation vf = vi + at.

## 5. Can final velocity be negative?

Yes, final velocity can be negative. This means that the object is moving in the negative direction of a coordinate system, and its speed is decreasing. In other words, the object is slowing down.

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