1. ## Big Problem :/

This problem is kinda big and just cant seem to start it off :/.

A new motion detector is used to graph the position of a sports car. If the themotion detector is not callibrated properly, it might give unrealistic graph data. The position of the car is described by the function S= 2t^3-21t^2 + 60, tE [0,6], where s is measured in meters and t in seconds. it is known the largest acceleration is 12 m/s ^2.

a) WHat is its initial velocity?
b)When is the car at rest?
c)When is the car at rest?
d)draw a diagram
e)Find the total distance traveled by the car in the first 6 seconds
f)Find the max acceleration of the car?

and i did a problem that i thought was to easy to be right just wondering if someone can second my answer, i doubt its right.

A stone is tossed down at 8 m/s from a edge of a cliff 63 m high how long will it take to hit the ground at the foot of the cliff?

3.8 seconds is what i got :/

Thanks for the help TJ

2. Originally Posted by tnkfub
This problem is kinda big and just cant seem to start it off :/.

A new motion detector is used to graph the position of a sports car. If the themotion detector is not callibrated properly, it might give unrealistic graph data. The position of the car is described by the function S= 2t^3-21t^2 + 60, tE [0,6], where s is measured in meters and t in seconds. it is known the largest acceleration is 12 m/s ^2.

a) WHat is its initial velocity?
b)When is the car at rest?
c)When is the car at rest?
d)draw a diagram
e)Find the total distance traveled by the car in the first 6 seconds
f)Find the max acceleration of the car?
So you have $\displaystyle s=2t^3-21t^2+60,0\leq t\leq 6$
To find its initial velocity is when $\displaystyle t=0$, but not in the formula above because that is for distance, but in its derivative (because derivative of distance is speed),
$\displaystyle v=6t^2-42t$ thus the initial velocity is zero.
The car is at rest when it have no speed, thus, $\displaystyle v=0$ solving $\displaystyle 6t^2-42t=0$ we factor $\displaystyle 6t(t-7)=0$ only solution is $\displaystyle t=0$ because $\displaystyle 0\leq t\leq 6$.
The total distance traveled is $\displaystyle \int^6_0|6t^2-42t|dt$ which is, 324.
The acceleration is the derivative of velocity thus,
$\displaystyle 12t-42$, it is maximum when $\displaystyle t$ is as large as possible, that happens when $\displaystyle t=6$ and its acceleration is 30.