Question

the position of a point during the interval of time from t=0 to t=6sec is given by S=-0.5t^3+6t^2+4t. at what time does the maximum level of the point occur. what is the maximum velocity and the acceleration when the velocity is maximum

Answer 1

@sweetburger @satellite73 @mathstudent55 @mathmath333 @

Answer 2

\[S=-0.5t^3+6t^2+4t\] right?

Answer 3

yup

Answer 4

at what time does the maximum level of the point occur?
take the derivative, set it equal to zero and solve

Answer 5

i sure hope this is a calc class, otherwise i am clueless

Answer 6

yah can you help me to aarrive for the right answer

Answer 7

sure
the derivative is a quadratic so it can't be that hard to solve

Answer 8

actually this function might be increasing on the interval from 0 to 6, largest at the endpoint
did you get the deivative

Answer 9

\[\frac{ dx }{ dt } = -1.5t^{2} +12 t +4\]

Answer 10

is tht right?

Answer 11

yeah and i think the zeros of this are outside your interval
lets check i t

Answer 12

how can you show to me?

Answer 13

\[-1.5t^2+12t+4=0\] i would get rid of the decimal and use the quadratic formula
maybe multiply by -2 and start with \[3t^2-24t-8=0\]

Answer 14

then?

Answer 15

\[t=\frac{24\pm\sqrt{12^2-4\times 3\times (-8)}}{6}\]

Answer 16

aka quadratic formula
and a calculator too for this
pretty sure the answer is 8 point something, outside your interval, so function has a max a at the right hand endpoint, namely at 6

Answer 17

as a matter of fact, i am sure of it
your function is increasing on the interval \([0,6]\) so it is largest at the right hand endpoint of the interval

Answer 18

ok that is wrong sorry
max velocity is the max of the derivative, aka the vertex

Answer 19

find the vertex of \[s'(t)=-1.5t^2+12t+4\] using \(-\frac{b}{2a}\)

Answer 20

to find the acceleration at that time, use the second derivative and plug in the first coordinate of the vertex of the first derivative

Answer 21

let me know if i lost you anywheres there