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Kinematics question help

  • #1

Homework Statement


A commercial claims that a car can stop on a dime (17.9mm). If a car is initially traveling at 13.4m/s, what acceleration would the car experience coming to a stop in that short distance?

Homework Equations


avg acceleration
Coversion

The Attempt at a Solution


I recognize that the acceleration may be negative since the car is coming to a stop. So i kept that in mind as I tried solving.

I first converted 17.9mm to m and got 0.179m.

I then tried to draw it out and label values.
  • Initial velocity: 13.4m/s
  • Acceleration: ?
  • Displacement: 0.179m
  • I'm assuming that the final velocity is 0m/s since its coming to a stop.
I then tried to solve for the time. But would that be irrelevant in this question?

The question is also multiple choice.
A) -5m/s^2
B) 250m/s^2
C) 10m/s^2
D) -5000m/s^2
E) 240m/s^2

So I'm thinking that the answer maybe either be A or D. I'm just not sure what steps to take next to get to either conclusion.
 

Answers and Replies

  • #2
Bystander
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  • #3
kuruman
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Homework Statement


A commercial claims that a car can stop on a dime (17.9mm). If a car is initially traveling at 13.4m/s, what acceleration would the car experience coming to a stop in that short distance?

Homework Equations


avg acceleration
Coversion

The Attempt at a Solution


I recognize that the acceleration may be negative since the car is coming to a stop. So i kept that in mind as I tried solving.

I first converted 17.9mm to m and got 0.179m.

I then tried to draw it out and label values.
  • Initial velocity: 13.4m/s
  • Acceleration: ?
  • Displacement: 0.179m
  • I'm assuming that the final velocity is 0m/s since its coming to a stop.
I then tried to solve for the time. But would that be irrelevant in this question?

The question is also multiple choice.
A) -5m/s^2
B) 250m/s^2
C) 10m/s^2
D) -5000m/s^2
E) 240m/s^2

So I'm thinking that the answer maybe either be A or D. I'm just not sure what steps to take next to get to either conclusion.
Solve what for time? Can you post what you did? Also, some relevant equations will help.
 
  • #4
  • #5
Solve what for time? Can you post what you did? Also, some relevant equations will help.
I look to 4 kinematic equations including one for average velocity and one for average acceleration (Vf: Final Velocity, Vi: Initial Velocity, a=Acceleration, t= Time, Δd=Displacemet); Δx: Change in position, Δt: Change in time
  • Vf=Vi+ a⋅t
  • Δd=Vi(t)+1/2a⋅t
  • Vf^2=Vi^2+2a⋅Δd
  • Δd=1/2(Vf+Vi)t
Average Velocity= Δx/Δt
Average Acceleration= Δv/Δt

I basically just label values and try to see which equation is best to solve. in this instance i used:

Vf^2=Vi^2+2a⋅Δd

Since i had displacement, initial velocity, and final velocity. I solved for a. My answer was -5015m/s^2
 
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  • #6
The initial velocity is given: its magnitude is 13.4m/s.
The final velocity has a magnitude of 0m/s since the car is no longer moving.
The displacement is given: its magnitude is 17.9mm (or 0.0179m).
Therefore, to find the acceleration, the following equation can be used:
Vf2 - Vi2 = 2ad
(0)2 - (13.4)2 = 2a(0.0179)
0 - 179.56 = 0.0358a
-179.56 = 0.0358a
a = - 5015.64 m/s2
The answer is closest to D, which must be the correct answer.
 
  • #7
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The initial velocity is given: its magnitude is 13.4m/s.
The final velocity has a magnitude of 0m/s since the car is no longer moving.
The displacement is given: its magnitude is 17.9mm (or 0.0179m).
Therefore, to find the acceleration, the following equation can be used:
Vf2 - Vi2 = 2ad
(0)2 - (13.4)2 = 2a(0.0179)
0 - 179.56 = 0.0358a
-179.56 = 0.0358a
a = - 5015.64 m/s2
The answer is closest to D, which must be the correct answer.
Posting full solutions is against the forum's rules (Short Summary → Point 3).
Instead, guide the seeker on the right track by allowing them to explore all aspects of a particular problem.
No offence to you, but the purpose of PF is to help the seeker understand the different sciences; and not to serve as a platform for showing off how bright the responder is.
 
  • #8
PeterO
Homework Helper
2,425
46

Homework Statement


A commercial claims that a car can stop on a dime (17.9mm). If a car is initially traveling at 13.4m/s, what acceleration would the car experience coming to a stop in that short distance?

Homework Equations


avg acceleration
Coversion

The Attempt at a Solution


I recognize that the acceleration may be negative since the car is coming to a stop. So i kept that in mind as I tried solving.

I first converted 17.9mm to m and got 0.179m.

I then tried to draw it out and label values.
  • Initial velocity: 13.4m/s
  • Acceleration: ?
  • Displacement: 0.179m
  • I'm assuming that the final velocity is 0m/s since its coming to a stop.
I then tried to solve for the time. But would that be irrelevant in this question?

The question is also multiple choice.
A) -5m/s^2
B) 250m/s^2
C) 10m/s^2
D) -5000m/s^2
E) 240m/s^2

So I'm thinking that the answer maybe either be A or D. I'm just not sure what steps to take next to get to either conclusion.
Note: When you have a multiple choice question, you are not really finding THE answer, you are identifying which of the offerings IS the answer.
I like that you knew acceleration would be negative, and thus eliminated 3 of the options.
If the answer was -5, then it would take between 2 and 3 seconds to stop from a speed of 13.4 m/s.
Given the average speed during the "slow down" is 6.7 m/s [(13.4 + 0)/2], in 2 seconds the car would travel 13.4 m, so in this case more than 13m.
This -5 can be excluded leaving just one option.

If you use the equations correctly, it will identify the answer correctly - just take longer. Use you time on the more difficult questions.
 
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