Word Problem: car is travelling at a rate of 66 feet per sec

[AMaccy]

Hi there,

I need some help with this problem:

"A car is travelling at a rate of 66 feet per second (45 miles per hour) when the brakes are applied. The position function for the car is given by s=-8.25t^2+66t, where s is measured in feet and t is measured in seconds. Create a table showing the position, velocity, and acceleration for each given value of t. What can be concluded?"

I know that :

v(t) = s'(t)

a(t) = v'(t) = s''(t)

However I can't plug in the math (I get lost) -- I was hoping someone can show me a step by step solution to this problem? I would really appreciate it.


Thanks!!

 

[tkhunny]

Find t such that s(t) = 0. What is the significance of this t? This defines the necessary Domain. Why?

 

[fasteddie65]

t x v
0 57.75 49.5
1 99 33
2 123.75 16.5
3 132 0

At t = 3 the car has stopped.

 

[BigGlenntheHeavy]

\(\displaystyle S(t) \ = \ -8.25t^{2}+66t\)

\(\displaystyle v(t) \ = \ -16.5t+66\)

\(\displaystyle a(t) \ = \ -16.5\)

\(\displaystyle The \ car \ stops \ when \ the \ velocity \ is \ zero, \ hence \ v(t) \ = \ 0 \ = \ -16.5t+66 \ implies \ t \ = \ 4sec.\)

\(\displaystyle S(4) \ = \ -8.25(16)+66(4) \ = \ 132 ft.\)

\(\displaystyle Ergo, \ when \ the \ brakes \ are \ put \ on \ the \ car \ goes \ 132ft \ in \ 4sec. \ as \ it \ decelerates \ 16.5ft/sec^{2}.\)