Acceleration Problems: Calculating Acceleration for Slowing Cars

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Homework Help Overview

The discussion revolves around calculating acceleration for cars slowing down under different conditions. The original poster presents two problems involving uniform deceleration and seeks clarification on the correct approach to find acceleration and distance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of kinematic equations to relate acceleration, initial and final velocities, and distance. There are attempts to clarify the concept of average velocity and its application in the problems. Questions arise regarding the correct interpretation of the equations and the algebra involved.

Discussion Status

Some participants provide guidance on using average velocity and kinematic equations, while others express confusion about specific calculations and the application of formulas. There is an ongoing exploration of different interpretations and methods without a clear consensus on the solutions.

Contextual Notes

Participants note the absence of time in the first problem, which complicates the calculations. There are also discussions about the correct application of algebra in manipulating the kinematic equations.

magdalenagirl
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Homework Statement


I) A car slows down from 23 m/s to rest in a distance of 80 m. What was its acceleration, assumed constant?

II) A car slows down uniformly from a speed of 26.0 m/s to rest in 5.50 s. How far did it travel in that time?

Homework Equations



what am i doing wrooong? am i using the right formula?

The Attempt at a Solution


I) I used this formula: a=Vf-Vi/tf-ti but the problem doesn't give me time so it didn't work

II) I used the same formula ^^ 0-26.0/0-5.50=4.72 is not the answer
 
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Well, you can find the time it took to travel 80m using Vavg and then find the acceleration. Or v^2 = vo^2 + 2ax
For the second part, you need to find distance not acceleration. Which is equal to Vavg*t
 
what's Vavg?
 
magdalenagirl said:
what's Vavg?

(v)avg = (v + vo)/2
 
Ok, I got problem # II.

I still don't understand # I...
So I find the Vavg = (0+23)/2 = 11.5
am I right here?
Then I guess I plug that into the formula:
11.5^2=0^2+2 a (80?) >> I don't know what's my X

Help!
 
Since the time is not given, use the kinematic equation that relates acceleration, displacement and the squares of the final and initial velocity.
 
kuruman said:
Since the time is not given, use the kinematic equation that relates acceleration, displacement and the squares of the final and initial velocity.

that's what I'm doing and I'm not getting the right answer
v^2 = vo^2 + 2ax

0=23^2+2a(80)
0=689a
 
magdalenagirl said:
that's what I'm doing and I'm not getting the right answer
v^2 = vo^2 + 2ax

0=23^2+2a(80)
0=689a

How did you combine those factors? The first does not have an a in it so cannot simply be added.
 
Your algebra is wrong.

0 = 232+2*a*80 is OK. It's the next step that is incorrect. Doing the number multiplications gives

0 = 529 + 160*a

You can't add 529 and 160*a to get 689*a ! To get 689*a you need to add 529*a + 160*a, which is not what you have here.

Instead, you say

160a = -529

a = -529/160 m/s2
 
  • #10
ok,
0(final)=23^2(initial) + 2 a (80)
I got the 23 ^2 which is 529
then I multiplied 80 * 2 =160
then I added 529 +160 =689
 
  • #11
oooooooh ok,
thanks a lot kuruman! I really appreciate it.
 

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