Question

@Jhannybean @ganeshie8 can you guys please explain this calc problem?!

Answer 1

just number 5!!

Answer 2

calculator problem ???

Answer 3

calculus darling :P

Answer 4

ohh okay
ik

Answer 5

sure u do ;)

Answer 6

where are you stuck ? just integrate the absolute value of velocity for the distance

Answer 7

\[\int\limits_0^8 |v(t)|~dt\]

Answer 8

First find the position function s(t)by integrating the velocity function as @ganeshie8 has suggested. Then the average velocity on the given interval will be found by\[\frac{ s(8)-s(0) }{ 8-0}\]

Answer 9

integrating: t^3/3-9t^t/2+18t+C , what do i do next?

Answer 10

#1 should make you think about average value of a funciton
thats precicely the reason we define average value of f(x) in interval [a, b] as
\[\large \dfrac{\int\limits_a^b f(x)~dx}{b-a}\]

Answer 11

yrah i did those, i only havnt done 5

Answer 12

just 5!

Answer 13

for #5, start by finding the absolute value of v(t)

Answer 14

set v(t) equal to 0 and solve t

Answer 15

x=6 and x=3

Answer 16

Excellent! next split the integral and work each of them

Answer 17

like plug them in?

Answer 18

i dont get what that means:/

Answer 19

What he means his now find the distance from t = 0 to t = 3, then from t = 3 to t = 6, then from t = 6 to t = 8. Then the sum of all these distances will be the total distance travelled from t = 0 to t = 8.

Answer 20

Notice v(t) is positive everywhere except in the interval [3, 6]
your goal is to integrate |v(t)|, so you may split the integral into 3 pieces such that the v(t) doesn't change sign in each interval:

\[\int\limits_0^8 |v(t)|~dt = \left|\int\limits_0^3 v(t) ~dt \right|+ \left|\int\limits_3^6 v(t) ~dt \right|+ \left|\int\limits_6^8 v(t) ~dt \right| \]

Answer 21

and solve each one?

Answer 22

yes, you may use the antiderivative that you have found earlier and simply work the bounds if you want

Answer 23

thank you(:

Answer 24

you're welcome!

Answer 25

If you already have the position function, which is\[s(t) = \frac{ 1 }{ 3 }t ^{3}-\frac{ 9 }{ 2 }t ^{2}+18t +1\]Then for example the distance from t = 0 to t = 3 is\[|s(3)-s(0)|\]