Question

Calculus help !

Answer 1

# 6 and 7# please. Not sure if Im doing it right

Answer 2

for 6a. and 6b.
use this :
\(\large Avg.~Velocity = \dfrac{total ~ distance }{total ~ time}\)

for 6c.
\(velocity =\dfrac{d}{dt}(distance) = s'(t)\)
that is differentiate s(t) and then plug in t=2

Answer 3

For 7 a.
rate of change in amount M is just \(\Large \dfrac{dM}{dt} \)
so differentiate M and plug in t=2

7b.
Positive answer means amount increases with time.
so, negative answer means....

Answer 4

Let me elaborate 6a. and 6b.

6 a. average velocity at 3rd sec. = distance traveled in 3 sec /3 sec
= s(3)/3

6b. average vel. in the interval (a,b) is
\(\Large = \dfrac{f(b)-f(a)}{b-a}\)

Answer 5

Need help with 6 c. I bit stuck. For 6b they got -10t m/s but not sure how. But before that step i got -55m/s for average velocity