Finding displacement through velocity and deceleration

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Homework Help Overview

The discussion revolves around a physics problem involving a car's displacement while decelerating. The car starts with an initial velocity of 10.0 m/s and decelerates at a rate of -2.00 m/s², reaching a final velocity of 5.00 m/s. Participants are exploring how to calculate the displacement using the appropriate kinematic equation.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correct application of the kinematic equation for displacement and express confusion regarding the signs and units involved in the calculations. There are attempts to clarify the meaning of negative acceleration and how it relates to deceleration.

Discussion Status

The discussion is ongoing, with participants providing guidance on the equation to use and questioning the handling of signs in the calculations. There is an exploration of unit cancellation and dimensional analysis, but no consensus has been reached on the final answer or the correct interpretation of the acceleration's sign.

Contextual Notes

Some participants note that the wording of the problem may lead to confusion regarding the signs of acceleration, and there is a focus on ensuring that the equation remains dimensionally correct throughout the discussion.

Grace204
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Homework Statement


a car with an initial velocity of 10.0 m/s (W) decelerates at -2.00 m/s (s to the power of 2) (W) reaching a final velocity of 5.00 m/s (W). What is the displacement of the car?

Homework Equations


Δdisplacement = (final velocity to the power of 2 - initial velocity to the power of 2)/ 2 times acceleration

The Attempt at a Solution


I didn't really get far because I didn't know how to deal with the units and the minus/plus signs
 
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Grace204 said:
decelerates at -2.00 m/s
I believe it should rather be 'accelerates at -2m/s2'. Negative acceleration is deceleration and negative deceleration is acceleration..
 
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true, I didnt know how to add the subscript on top. But can you help me answer this?
 
Your relevant equation is correct. Put the known values in that equation and solve for displacement.
s=(v2-u2)/2a
 
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cnh1995 said:
Your relevant equation is correct. Put the known values in that equation and solve for displacement.
s=(v2-u2)/2a
I don't know how to. The anwer I am getting is -18.75, and i don't think that's right.
 
Grace204 said:
I don't know how to. The anwer I am getting is -18.75, and i don't think that's right.
It is negative because you used a=2m/s2
Since the car is decelerating, you should use a=-2m/s2..
Then you'll get it as 18.75m.
 
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This is what i did-
(5.00 m/s [W])2 - (10.0 m/s [W])2 / 2 * -(-2.00 m/s2
(-5.00 m/s)2 - (-10.0 m/s)2 / 4.00 m/s2
After this, what happens to the units? How do the s2 cancel each other out?
 
Grace204 said:
This is what i did-
(5.00 m/s [W])2 - (10.0 m/s [W])2 / 2 * -(-2.00 m/s2
(-5.00 m/s)2 - (-10.0 m/s)2 / 4.00 m/s2
After this, what happens to the units? How do the s2 cancel each other out?

Do the units become m2/s2 divided by m/s2?
 
Grace204 said:
This is what i did-
(5.00 m/s [W])2 - (10.0 m/s [W])2 / 2 * -(-2.00 m/s2
(-5.00 m/s)2 - (-10.0 m/s)2 / 4.00 m/s2
After this, what happens to the units? How do the s2 cancel each other out?
Yes. And only the unit of displacement remains which is meter..Every equation has to be dimensionally correct.
 
  • #10
cnh1995 said:
Yes. And only the unit of displacement remains which is meter..Every equation has to be dimensionally correct.
So s2 divided by s2 cancels each other out and does not equal s4? Because acceleration is m/s divided by s
 
  • #11
Grace204 said:
This is what i did-
(5.00 m/s [W])2 - (10.0 m/s [W])2 / 2 * -(-2.00 m/s2
(-5.00 m/s)2 - (-10.0 m/s)2 / 4.00 m/s2
After this, what happens to the units? How do the s2 cancel each other out?
Why did you put double negative signs on the acceleration in the denominator?

You should be able to work out the resulting units using standard algebra.

If you square velocity in m/s, what do you get?

If you divide these units for velocity squared by the units for acceleration, what's left?
 
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  • #12
Grace204 said:
m2/s2 divided by m/s2?
Yes. You can see s2 cancel each other and m2/m becomes m.
 
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  • #13
SteamKing said:
Why did you put double negative signs on the acceleration in the denominator?

You should be able to work out the resulting units using standard algebra.

If you square velocity in m/s, what do you get?

If you divide these units for velocity squared by the units for acceleration, what's left?

I put double negatives on the denominator because the question says is "decelerates at -2.00 m/s2 [W] or -(-2.00 m/s2)"
 
  • #14
Grace204 said:
I put double negatives on the denominator because the question says is "decelerates at -2.00 m/s2 [W] or -(-2.00 m/s2)"
Is this wrong? If so, how would I write this out?
 
  • #15
Grace204 said:
Is this wrong? If so, how would I write this out?
The question is poorly worded. One negative sign should suffice, since 'decelerate' means to reduce speed or velocity.
 
  • #16
Grace204 said:
Is this wrong? If so, how would I write this out?
The initial and final velocities are 10m/s and 5m/s respectively. This clearly tells you that the car is 'decelerating'. So, you should write either 'a car is decelerating at 2m/s2' or ' a car is accelerating at -2m/s2'. That's what I meant in #2.
You should write the equation in the following form.
cnh1995 said:
s=(v2-u2)/2a
Now substitute a=-2 here, since the car is decelerating.
 
  • #17
Thank you so much for your help!
 
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