Distance Traveled During 5th Second of Coasting

AI Thread Summary
The discussion revolves around calculating the distance a car travels during the fifth second of coasting after decelerating from 33.5 km/hr at a rate of 0.29 m/s². Participants clarify that the distance during the fifth second refers to the interval from 4 seconds to just before 5 seconds, not the total distance covered in the first five seconds. The correct approach involves using kinematic equations to find the displacement at the specified time intervals. One participant initially miscalculates the distance but later acknowledges the misunderstanding and corrects their approach. The conversation emphasizes the importance of accurately interpreting time intervals in kinematic problems.
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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 \Deltax 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 \Deltax 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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