Question

Will fan and Medal!!

I just need help on problems 41, 49, 52, and 58. I've done all the rest! Click for link to problems

Answer 1

Please, even if you can only help with one! I can always just post another question for the rest

Answer 2

41. The book hits the ground at p(t) = 0. Solve for t.
Also, v(t) = p'(t). Use this to find the speed when p(t) = 0.

Answer 3

49. Seems like part A is asking you to pick points off the graph and use them to find 2 estimates of the area under the curve from 0 to 4. (left, right, midpoint, trapezoid, etc)

B. \[\int\limits_{0}^{4}a(c)~dc\]

Answer 4

52. If you assume forward is the positive direction, then the car moves forward when v(t) > 0, backward when v(t) < 0 and is not moving when v(t) = 0.

Answer 5

58. that integral and R(T) tell you the same thing, the barrels of oil pumped in T days. So R(5) gives how many barrels were pumped in 5 days.

Answer 6

ok i understand all of your explanations except for 41.

Answer 7

@peachpi am I solving for t when p(t)=0 or am I solving for t for the equation
p(t)=1250-16t^2?

Answer 8

you're setting p(t) equal to 0 then solving for t
p(t) = 1250 - 16t^2
p(t) = 0

1250 - 16t^2 = 0

Answer 9

@peachpi x=8.84??

Answer 10

yes

Answer 11

that's the time when it hits the ground. Now integrate to find the speed at this time

Answer 12

find the derivative of 8.84? or the equation?

Answer 13

the equation

Answer 14

do i plug in 8.84 for t?

Answer 15

yes

Answer 16

and that should give you velocity. I think you can just make velocity positive to get speed

Answer 17

-.3296

Answer 18

and then make it positive: .33

Answer 19

so velocity is about .33m/s?

Answer 20

and that would be the exact speed for when the book hits the ground?

Answer 21

way too low. what was your derivative?

Answer 22

should have had v(t) = p'(t) = -32t.
then plug in 8.84 for t

Answer 23

OH woops I plugged in 8.84 for 1250-16(8.84)

Answer 24

ok i got what you did

Answer 25

and make it positive?

Answer 26

so v(t)=282.88m/s??

Answer 27

yes

Answer 28

hooraaay! what about for a, where i need 3 approximations for speed using different size intervals? is that left and right hand sums or something else?

Answer 29

I think they just want you to find the average rate of change on 3 different intervals. The smaller will be the most accurate because it's a smaller snapshot and should be closer to the instantaneous speed you just found.

Since you're investigating what happens at t = 8.84 I'd use these intervals
7 s to 8.84 s
8 s to 8.84 s
8.5 s to 8.84 s

Answer 30

ok so for example on the interval from 7 to 8.84 seconds just plug both of those numbers into the equation? like over and under estimates? and the most accurate one would be the third interval

Answer 31

but i don't need to take the derivative correct? or do i in order to get velocity?

Answer 32

no you don't need to take the derivative. Avg rate of change will give you an estimate for the velocity.

Basically you use this equation
\[aroc=\frac{f(b)-f(a)}{b-a}\]

Answer 33

so 8.84-7 divided by what?

Answer 34

For the first interval
p(7) = 466
p(8.84) = 0

\[\frac{p(8.84)-p(7)}{8.84-7}=\frac{0-466}{8.84-7}=-253.3\]

Then make 253.3 for speed

Answer 35

so the speed for that interval is 253.3m/s?

Answer 36

yes

Answer 37

you got it from here?

Answer 38

yes i think so ! thank you soooo much

Answer 39

you're welcome