Physics 11 Speed and acceleraton

[anOldMan]

[SOLVED] Physics 11 Speed and acceleraton

Homework Statement

Linda sees an elephant dart into the road 50.0m ahead of her car while she is driving at 65km/hr. She slams on her brakes, which decelerate the car at the rate of -5.80m/s^2. Will she be able to avoid hitting the elephant? Find her stopping distance.

Answer is Yes, stopping distance is 28.1

Homework Equations

The Attempt at a Solution

65km/hr = 18.056m/s $$V_{avg}9.028$$
a = v / t -5.80m/s^2 = 9.028m/s / T = 1.56 seconds
using $$d = v_{0}t+.5at^{2}$$
d = .5(-5.80m/s^2) (2.43s^2) = 7meters
I am missing something here.

Thanks for the help on the previous question and thanks for any help for this question. Your quick response is appreciated.

 

[anOldMan]

I seem to have found the answer, but why is it that the rate is not averaged? The rate is going down from 65 to 0. Therefore, the rate should be an average before doing the calculations. I am confused.

 

[physixguru]

V=0
U=18M/S
A=-5.80 M/S^2

V^2=U^2+2AS
>0=324-11.6S
>S=27m

why would you average the rate?since the deceleration is uniform..u must not average
the rate.this shall lead to loss in generality of the problem.

 

[Hootenanny]

V=0
U=18M/S
A=-5.80 M/S^2

V^2=U^2+2AS
>0=324-11.6S
>S=27m

Careful with rounding errors.

 

[SheldonG]

I seem to have found the answer, but why is it that the rate is not averaged? The rate is going down from 65 to 0. Therefore, the rate should be an average before doing the calculations. I am confused.

By rate, do you mean acceleration? For the case of constant acceleration, average acceleration = instantaneous acceleration.

 

[anOldMan]

Thanks guys.

why would you average the rate?since the deceleration is uniform..u must not average
the rate.this shall lead to loss in generality of the problem.

This solved my question. I thought about it again and realized that I made an error. In any case, it will take me a while to wrap my mind around these questions. The deceleration is uniform and will go from 65 km/h to 0 km/h in a certain amount of time based on the rate of deceleration. The reason I was averaging the number was because of the formula sheet. It is written as
note: (change) = a triangle. I assume the triangle in front of the variable means change.
a=(change)v/(change)t

I assumed the change meant average.