Positive or negative final velocity? Constant acceleration

[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.

 

[MarkFL]

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.

 

[DrClaude]

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 $v_f = \pm 2 \mathrm{m/s}$.

by doing a velocity vs time graph it seems that +2 m/s is the right answer.

Can you post that graph?

 

[cnh1995]

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

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.

 

[AnnaRiddle]

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.

 

[Chestermiller]

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

 

[AnnaRiddle]

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.

 

[Chestermiller]

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?

 

[Ray Vickson]

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.