Distance Traveled During 5th Second of Coasting

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Homework Help Overview

The problem involves a driver slowing a car from an initial speed of 33.5 km/hr at a constant deceleration of 0.29 m/s². The objective is to calculate the distance the car travels specifically during the fifth second of coasting.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the correct interpretation of the time interval for the fifth second, with some confusion about whether it refers to the time from 4 to 5 seconds or from 5 to 6 seconds. There are attempts to apply kinematic equations to find displacement, but misunderstandings about the question's requirements are noted.

Discussion Status

Some participants have offered guidance on using kinematic equations and clarified the time intervals involved. There is acknowledgment of confusion regarding the definition of the fifth second, but no consensus has been reached on the correct approach yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The original poster's calculations have been questioned, indicating a need for further exploration of the problem's requirements.

black_hole
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Homework Statement


A driver slows her car from 33.5 km/hr at a constant rate of 0.29 m/s2 just by taking her foot of the accelerator (the "gas" pedal). Calculate the distance the car travels during the fifth second that she is coasting.


Homework Equations





The Attempt at a Solution


I am doing something wrong b/c I get 42.905m, but I know that is wrong.
 
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How are you arriving at that answer? If you post your steps then someone can show you where you went wrong.
 
Use your kinematics.

You know the following:

vi = 33.5 km/hr
a = -0.29 m/s2

"During the fifth second" means you start at 4 seconds and end an instant before the 5th second. So, you need to find the [tex]\Delta[/tex]x at t = 4 and t = 5.

Now, what you do with that knowledge is up to you. I probably told you too much as it is.

Good luck!
 
Last edited:
initial velocity = 9.306m/s
acceleration = -.29m/s/s
time = 5s
displacement = x

x = initial velocity *time + 1/2at*t
x = 42.905m
 
black_hole said:
initial velocity = 9.306m/s
acceleration = -.29m/s/s
time = 5s
displacement = x

x = initial velocity *time + 1/2at*t
x = 42.905m

Okay, all you found was how far it went in the FIRST FIVE seconds. The question (at least the way your worded it) wants to know how far it goes DURING THE FIFTH second. Do you see the difference? Refer to my previous post if you are confused. If you are still confused after reading that, let us know!
 
JDHalfrack said:
Use your kinematics.

You know the following:

vi = 33.5 km/hr
a = -0.29 m/s2

"During the fifth second" means you start at 5 seconds and end an instant before the 6th second. So, you need to find the [tex]\Delta[/tex]x at t = 5 and t = 6.

Now, what you do with that knowledge is up to you. I probably told you too much as it is.

Good luck!

Would that not be during the 6th second? From 0s to 1s would be the first second, 1s to 2s is the second second etc., So the 5th second should be from 4s to 5s?
 
Would that not be during the 6th second? From 0s to 1s would be the first second, 1s to 2s is the second second etc., So the 5th second should be from 4s to 5s?

Yeah, I'm an idiot. Thank you. The idea was right, I just confused myself. Whoops!

Thanks for the correction. I fixed it.
 

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