Solving distance in this problem. Easy

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Homework Help Overview

The discussion revolves around a physics problem involving the braking distance of a car that decelerates from an initial velocity of 96 km/h to rest over a period of 3.0 seconds. The participants are exploring the calculations related to deceleration and the distance traveled during braking.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of deceleration and its implications for determining the distance traveled. There are attempts to apply equations of motion, with some questioning the correctness of the units used in the calculations. Others suggest alternative equations for solving the problem.

Discussion Status

The discussion is ongoing, with participants providing feedback on each other's calculations and questioning the validity of the units. Some guidance is offered regarding the use of different equations of motion, indicating a productive exploration of the topic.

Contextual Notes

There are indications of confusion regarding the units of deceleration, with a specific mention that it should be expressed in m/s². The original poster's repeated question about the distance traveled suggests a focus on clarifying the problem setup.

kencamarador
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During the braking test of. 1988 car, the car came to rest from an initial velocity of 96km/h (w) in 3.0 seconds. Assuming that deceleration remains constant.

How far did the car travel after brakes were applied?

I calculated the deceleration. It is 8.9m/s (e)
 
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kencamarador said:
During the braking test of. 1988 car, the car came to rest from an initial velocity of 96km/h (w) in 3.0 seconds. Assuming that deceleration remains constant.

How far did the car travel after brakes were applied?

I calculated the deceleration. It is 8.9m/s (e)

Keep going...
 
I used this equation to solve for Distance.

Vf^2 - vi^2/ 2 aav

VHF is final velocity and vi is initial velocity, aav is average velocity

=0^2 - 96^2/ 2 (8.9)

=-517.75

...
 
berkeman said:
Keep going...

There
 
kencamarador said:
I used this equation to solve for Distance.

Vf^2 - vi^2/ 2 aav
This equation is missing something.[/QUOTE]

One thing is an equal sign (so it's not an equation).

It's missing something else also.
VHF is final velocity and vi is initial velocity, aav is average velocity

=0^2 - 96^2/ 2 (8.9)

=-517.75

...
 
kencamarador said:
During the braking test of. 1988 car, the car came to rest from an initial velocity of 96km/h (w) in 3.0 seconds. Assuming that deceleration remains constant.

How far did the car travel after brakes were applied?

I calculated the deceleration. It is 8.9m/s (e)

Are you sure? This seems wrong to me (units?)
 
sjb-2812 said:
Are you sure? This seems wrong to me (units?)

8.9 is the deceleration.

My question is

How far did the car travel after brakes were applied?
 
kencamarador said:
8.9 is the deceleration.

Nope. As sjb-2812 has suggested the units are wrong. It should be 8.9 m/s2

My question is

How far did the car travel after brakes were applied?

I would use a different equation of motion...

http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

eg..

V2 = U2 + 2as

so

s = (V2 - U2)/2a
 

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