Question

A ball moves in a straight line and has an acceleration of a(t) = 2t + 5. Find the position function of the ball if its initial velocity is -3 cm per second and its initial position is 12 cm.

Answer 1

ok; tell me what you get when you integrate a(t) please

Answer 2

just t ?

Answer 3

i think you confusing the issue with derivatives...

try to think about what the function would look like that derives down to a(t).

in other words:

f(x) = ??? such that;

f'(x) = 2t +5

Answer 4

F(t)=2t+5

Answer 5

lets test that:

F(t) = 2t+5 ; when we derive it we get...

F'(t) = 2 then right? so that wont work

Answer 6

think of integrateing as "undoing" a derivative

Answer 7

the power rule for derivatives is:

X^n --> nX^(n-1); in this case n=2 right?

Answer 8

so it comes from something that resembles t^2 at least. so lets test this option and see where it gets us.

t^2 derives to .... 2t. its a perfect match :)

Answer 9

now we focus on the "5". How do we get a derivative that is 5? its just a constant sitting there without a variable ...

Answer 10

isn't the variable 12?

Answer 11

i dont think that 12 can be a variable; many people would confuse it to be the number 12 so that wouldnt be a good option in my book :)

Answer 12

the constant rule for derivatives says that if K is a constant then:

Kx derives to: K; in this case K=5 so lets try 5t as its antiderivative

5t derives to 5, it fits :)

this brings us to v(t) = t^2 + 5t; plus some constant that we need to solve for to anchor this floater to the y axis

v(t) = t^2 +5t + C

Answer 13

the term "initial" is code for : when t=0, so we can solve this v(t) specifically by making t=0 and the value of v(0) = -3 as stated in the problem

Answer 14

-3 = (0)^2 + 5(0) + C
-3 = C

v(t) = t^2 +5t -3; and in order to get to the position function we antiderive this function

Answer 15

I hate! and absolutely hate anti derivatives!

Answer 16

t^2 comes from some for of t^3...but when we derive t^3 we get:

[t^3]' = 3t^2 .... we get a pesky 3 in the way, so we need to get rid of it

what number can we divide by to get rid of the 3? im gonna say "3" lol

(t^3)/3 might work for us, lets test it

[(t^3)/3]' = (3/3)t^2 = t^2 .... it works!! :)

Answer 17

the power rule for antiderivatives is simply:

X^n comes from [X^(n+1)]/(n+1)

notice the t^2 antiderives to: (t^3)/3 as a result

Answer 18

I thought we had to find s(t) ?

Answer 19

5t follows the same rule; the exponent being a "1"

\[\frac{5t^{1+1}}{(1+1)}=\frac{5}{2}t^2\]

Answer 20

we are..we are; but it doesnt just magically appear :)

Answer 21

So would it be s(t)= 3t^3+5t^2-3t+12 ???

Answer 22

now for that " -3"; the rule is simply, attach the variable to it.

-3 antiderives to -3t ... plain and simple

when we put this all together we get

v(t) = t^2 +5t -3 antiderives to:

s(t) = (t^3)/3 +(5/2)t^2 -3t +C

Answer 23

I think I narrowed it down to that. Idk if it's right

Answer 24

amistre help me ; ) plzzz

Answer 25

Amistre is my baby! leave us alone

Answer 26

and you got ahead of me but yes :)

at the initial condition, code for t=0, we need s(0) = 12

12 = 0+0+0+C
12 = C

s(t) = (t^3)/3 +(5/2)t^2 -3t + 12 :)

Answer 27

youre structure is off a bit i think

Answer 28

s(t)= 3t^3+5t^2-3t+12

is not

s(t) = (t^3)/3 +(5/2)t^2 -3t + 12

Answer 29

Yeah I found my mistake (:

Answer 30

omg I get it! kinda ! but I understood it :D

Answer 31

lol good :)

Answer 32

babe!
now help me pllz!!!