# Kinematics question help

## 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?

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.

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Bystander
Homework Helper
Gold Member
17.9mm to m and got 0.179m.
You're sure?

kuruman
Homework Helper
Gold Member

## 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?

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.

You're sure?
Damn. Right. its 0.0179m x.x

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

kuruman
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:
(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.

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:
(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.

PeterO
Homework Helper

## 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?

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.

Bystander