Question

s(t)=-6t^2+12t+24
find displacement over interval of [0,3]

Answer 1

also find total distance travelled

Answer 2

displacement:

(3,s(3))
-(0,s(0))
---------
3 ,s(3)-s(0) ; pythag

\[\sqrt{3^2 +(s(3)-s(0))^2 }\]
\[\sqrt{9 +(s(3)-s(0))^2 }\]

Answer 3

distance traveled:

\[\int_{0}^{3}\sqrt{1+(-6t^2+12t+24)'^2}dx\]
\[\int_{0}^{3}\sqrt{1+(-12t+12)^2}dx\]
\[\int_{0}^{3}\sqrt{1+(144t^2-288t +144)}dx\]
\[\int_{0}^{3}\sqrt{1+144t^2-288t +144}\ dx\]

Answer 4

im sure that simplifies inthe end :)

Answer 5

yes..thanks

Answer 6

i think you have to be careful here

Answer 7

although i think the answer is easy enough to get. you can see from this equation that you have a parabola that faces down. it has a maximum at x = 1

Answer 8

at
\[x=0\] you are at 24 and at
\[x=1\] you are at 30 so you have traveled a total distance of 6 then you turn around and go back

Answer 9

at
\[x=3\] you are at 6 if my arithmetic is correct. so from 30 to 6 is 24 and your total distance traveled is 6+ 24 = 30

Answer 10

i am assuming here that
\[s(t)\] is position. if it is speed, then i am totally wrong.

Answer 11

but i think if it is just speed, then since your speed is positive over this entire interval, you can just take
\[\int_0^3(-6t^2+12t+24)dt \] to get your answer