Question

Hi guys. Having trouble with acceleration. Easy stuff but no textbook or teacher. Question is: "A car driver brakes gently. Her car slows down from 23ms^-1 to 11ms^-1 in 20 seconds. Calculate the magnitude of her deceleration. Can anyone walk me through this?

Answer 1

Hi! I think I might might be able to help! No textbook and no teacher stinks!!

So, can you tell me what acceleration is, generally?

Answer 2

I know that acceleration is when your picking up speed or slowing down. Like when an object is at rest and begins to start moving, it accelerates. I also know the formula for acceleration is change in speed over change in time.

Answer 3

Those are both right!
So,
1. something is at rest and begins moving
2. change in velocity over change in time

So, acceleration is when the velocity changes over some time!

Now, that equation you mentioned has everything you need for this problem.

One way to write it down is
\(a=\dfrac{v_f-v_i}{\Delta t}\)
That is, the acceleration is equal to the change in velocity (final minus initial will find the change) divided by the change in time.

Can you find those quantities in the problem that you were given?

Answer 4

The final velocity would be 11ms^-1 (going to leave out the units from here on out and add them on at the end) and the initial velocity will be 23. the time it takes for the change is 20 seconds. So a= -12/20 or -6ms^-1

Answer 5

I think I am overthinking the question, I thought it wanted me to calculate how far of a distance she covers when decelerating since ti asked for magnitude of the deceleration. Is all they want for the answer -6?

Answer 6

Well, they do want the distance, but I pointed you towards the acceleration for a reason! We'll use another equation that needs the acceleration.

But first, let's recheck the acceleration!
You said we had
\(a=\dfrac{12}{20}\)
So, that's really \(\dfrac{12}{20}=\dfrac{6}{10}=\dfrac{3}{5}\)
And remember to bring in the units then!
\(\rm\dfrac{ms^{-1}-ms^{-1}}{s}=\rm\dfrac{ms^{-1}}{s}=ms^{-2}\)

Answer 7

Oh.. And there was that negative sign, woops!

Answer 8

so, what is our acceleration in the end, with the units?

Answer 9

Hmm, would it be -.6 meters per second?

Answer 10

And why does it go from ms^-1 to ms^-2? that threw me off on the question I did after this.

Answer 11

acceleration would actually be \(0.6\rm m/s\color{blue}{^2}\)

It's tricky, but you have to look at it like this:
\(\sf\dfrac{velocity}{time}=\rm\dfrac{m/s}{s}=\dfrac ms\dfrac1s=\dfrac {m}{s^2}\)
So acceleration, by its nature, will always have those units, "meters per second squared."

Answer 12

Wouldn't the acceleration be negative because its slowing down? The change in speed is final - initial and final is 11 and initial is 23.

Answer 13

Oh, you're right about the value! I forgot about that negative sign again!! Good one!

Answer 14

I meant to emphasize on its unit, meters per second squared.

Are you okay with that?

Answer 15

Yeah that makes sense to me now, another question was asking for the answer in ms^-2 so I was confused about whether I had to do it by milliseconds or what. That clears it up though. So is -.6 m/s are final answer or do we have to keep going?

Answer 16

If its asking the magnitude of the deceleration, the magnitude is just the number right

Answer 17

Oh, yeah, that's it! I said something about distance earlier, but I was thinking of another question. The magnitude is the whole \(.6\rm\ m/s^2\).

Acceleration is two things:
1. magnitude
2. direction
So, it is \(0.60\rm\ m/s^2\) \(\sf\color{green}{in\ some\ direction}\).
This question just wants that first part, the magnitude.

Answer 18

Gahh!!!! \(-0.60\rm\ m/s^2\)

Answer 19

negative... negative... negative...

Answer 20

So since it just wants the magnitude and not the direction do we leave it just positive?

Answer 21

:)

Answer 22

Oh ok :P

Answer 23

If you aren't too busy, would you mind helping me with one more? I did it but I think I did it wrong.

Answer 24

Okay! Why don't you post it as another question for the sake of organization and I'll check over what you do! :)

Answer 25

Alright! Ill start a new thread now.

Answer 26

Okay!