Question

help with a calc problem!!! please explain step by step!

Answer 1

you know that acceleration is \(\large\color{slate}{\rm s''(t)}\)

Answer 2

So, given that your acceleration is 10ft/sec, you would need to integrate \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}10~dt}\)

Answer 3

can you calculate this integral ?

Answer 4

10t+C

Answer 5

So, \(\large\color{slate}{\displaystyle s'(t)=10t+C}\), and we have to figure out what C is.

You are given that initial velocity is -20ft/sec.

NOW, initial velocity is at t=0, so you are pretty much having that \(\large\color{slate}{\displaystyle s'(0)=-20}\)

Answer 6

So what can you say about the value of C ?

Answer 7

well, s'(t)=10t+C

s'(0)=C

and you are given the value of s'(0) to be -20, so what is your C ?

Answer 8

s'(0)??

Answer 9

s'(0) = ?

Answer 10

0?

Answer 11

that is given by the problem, explicitly.
look what is the initial velocity ?

Answer 12

(and the initial velocity, don't forget, is s'(0) )

Answer 13

initial velocity is -20ft/sec isn't ?

Answer 14

yes

Answer 15

and you can now solve for s'(0) (saying, solve for the initial velocity in terms of C)

\(\large\color{slate}{ s'(t)=10t+C }\)
\(\large\color{slate}{ s'(0)=10(0)+C }\)
\(\large\color{slate}{ s'(0)=C }\)

so you know that:
\(\large\color{slate}{ s'(0)=C }\)

and the problem gives you that \(\large\color{slate}{ s'(0)=-20 }\), so what is the value of C ?

Answer 16

-20

Answer 17

yup

Answer 18

So you have the entire s'(t) now. \(\large\color{slate}{ s'(t)=10t-20 }\)

Answer 19

to find the position you will need to integrate the velocity function, and than given that initial position (i.e. s(0) ) is 8 -- to solve for C.

Answer 20

what is \(\large\color{slate}{ \displaystyle\int\limits_{~}^{~}10t-20~dt }\) ?

Answer 21

5t^2-20+C

Answer 22

a little error

Answer 23

just -20 ?

Answer 24

20t!

Answer 25

yes, don't say !, (because ! is a factorial ) I know what you mean though.

\(\large\color{slate}{ \displaystyle\int\limits_{~}^{~}10t-20~dt =5t^2-20t+C }\)

Answer 26

you are given that initial position is 8 ft, correct ?

Answer 27

hahaha oops

Answer 28

yes:)

Answer 29

yes

Answer 30

(lol)...
now, the initial position is s(t) evaluated at t=0, SAYING, it is s(0).

Answer 31

\(\large\color{slate}{ \displaystyle s(0)=8 }\)

Answer 32

\(\large\color{slate}{ \displaystyle s(t)=5t^2-20t+C }\)
set t=0, and solve for C.

Answer 33

if you set t=0,
\(\large\color{slate}{ \displaystyle s(t)=5t^2-20t+C }\)
\(\large\color{slate}{ \displaystyle s(0)=C }\)

so you know that
\(\large\color{slate}{ \displaystyle s(0)=C }\)
and that
\(\large\color{slate}{ \displaystyle s(0)=8 }\)

Answer 34

what is the C ?

Answer 35

8

Answer 36

now, what is the \(\large\color{slate}{ \displaystyle s(t) }\), can you write it out entirely ?

Answer 37

(I mean without the C)

Answer 38

i dont know..

Answer 39

\(\large\color{slate}{ \displaystyle s(t) =5t^2-20t+C }\) (is what you had)
you found that C=8, so what is the \(\large\color{slate}{ \displaystyle s(t) }\) ?

Answer 40

168?

Answer 41

5t^2-20t+8

Answer 42

oh haha

Answer 43

yes, so \(\large\color{slate}{ \displaystyle s(t)=5t^2-20t+8}\)

Answer 44

so thats the position function?

Answer 45

thanks!!