Acceleration Problems: Calculating Acceleration for Slowing Cars

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A car slowing from 23 m/s to rest over 80 m requires the use of the kinematic equation v^2 = vo^2 + 2ax to find acceleration, as time is not provided. The correct approach involves rearranging the equation to isolate acceleration, leading to a calculation of a = -529/160 m/s². For the second scenario, where a car decelerates from 26.0 m/s to rest in 5.50 seconds, the average velocity is calculated as Vavg = (0 + 26.0)/2. The distance traveled can then be determined using the formula distance = Vavg * time. Proper understanding of the kinematic equations is essential for solving these types of problems.
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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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