Solving distance in this problem. Easy

[kencamarador]

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)

 

[berkeman]

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...

 

[kencamarador]

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

...

 

[kencamarador]

Keep going...

There

 

[SammyS]

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

...

 

[sjb-2812]

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?)

 

[kencamarador]

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?

 

[CWatters]

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