Question

?

Answer 1

@SolomonZelman

Answer 2

Differentiate

Answer 3

The derivative of the position-function will give you the velocity.
Then, when you plug in t=c (that is, f'(c)) that would give you the
velocity at t=c.

Answer 4

whats the formula

Answer 5

\(\color{#000000}{ \displaystyle s(t)=4t^2-3 }\)

Is the position-function.

Answer 6

Right?

Answer 7

okay

Answer 8

\(\color{#000000}{ \displaystyle s(t)=4t^2-3 }\)

You need to take the derivative of s(t).
Do you know how to do this?

Answer 9

no explain please

Answer 10

I am just curious.
Is this indeed Calculus, and is s(t) indeed the position-function (not function for speed)?

Answer 11

yes is calc

Answer 12

and s(t) is the position, as I understand, correct?

Answer 13

yes

Answer 14

So, you have two rules. (a and m are constants, and \(m\ne0\).)

\(\color{#000000}{ \displaystyle [1]\quad \quad \frac{d}{dt}t^m=mt^{m-1} }\)

\(\color{#000000}{ \displaystyle [2]\quad \quad \frac{d}{dt} (a)=0 }\)

Answer 15

And from [1], you can infer [3],

\(\color{#000000}{ \displaystyle [3]\quad \quad \frac{d}{dt}at^m=amt^{m-1} }\)

(coefficient "a" doesn't change anything)

Answer 16

what do you plug in

Answer 17

So, if my s(t) was

\(\color{#000000}{ \displaystyle s(t)=5t^3+6t^2+7t+2 }\)

rewriting the function,
\(\color{#000000}{ \displaystyle s(t)=5t^3+6t^2+7t^1+2 }\)

Then the derivative (denoted as s'(t) or as ds/dt) is:
\(\color{#000000}{ \displaystyle s'(t)=(\color{blue}{3})5t^{\color{blue}{3}-\color{red}{1}}+(\color{blue}{2})6t^{\color{blue}{2}-\color{red}{1}}+(\color{blue}{1})7t^{\color{blue}{1}-\color{red}{1}}+0 }\)

Notice, I apply rule [2], to the "2" at the end, and I
am using the rule [3] for all other terms.

So, when I simplify this, I get:
\(\color{#000000}{ \displaystyle s'(t)=15t^2+12t^1+7t^0+0 }\)

You know that \(\color{#000000}{ x^0=1 }\) in general, so you write the third term,
\(\color{#000000}{ \displaystyle s'(t)=15t^2+12t^1+7(1)+0 }\)

And that simplifies completely, to,
\(\color{#000000}{ \displaystyle s'(t)=15t^2+12t+7 }\)

Then, if I wanted to find the instantaneous velocity at \(\color{#000000}{ \displaystyle t=2 }\)
I would simply plug that into the derivative s'(t).

\(\color{#000000}{ \displaystyle s'(\color{darkgoldenrod}{2})=15(\color{darkgoldenrod}{2})^2+12(\color{darkgoldenrod}{2})+7 }\)
\(\color{#000000}{ \displaystyle s'(2)=15(4)+24+7=60+31=91 }\)
(And that would be the answer, 91!)

Answer 18

THAT IS JUST AN EXAMPLE OF A PROBLEM SIMILAR TO YOURS.
(except that my function is harder to differentiate than your function.)

Answer 19

@SolomonZelman is mine t=37

Answer 20

you mean the velocity at t=5, not "t", right?

Answer 21

i got 37 after doing out the work

Answer 22

\(\color{#000000}{ \displaystyle s(t)=4t^2-3 }\)

\(\color{#000000}{ \displaystyle s'(t)=(2)4t^{2-1}+0 }\)

The derivative of "-3" is 0, using rule [2].
The derivative of 4t\(^2\) is given by rules [3].

\(\color{#000000}{ \displaystyle s'(t)=8t }\)

So, your velocity is not 37.
(You forgot to differentiate the -3, I guess...)

Answer 23

so 3 is d?

Answer 24

Apologize. I don't quite understand what you asked.

Answer 25

In rule 2, 3 is inplace of d

Answer 26

Well, in rule [2], you are just given the information that the derivative of any constant (a) is 0.

Answer 27

So, since -3 is also a constant, therefore its derivative is 0.

Answer 28

So, what I did is differentiated the function s(t) by taking the derivative of each term separately, and this gave me the s'(t) (or the velocity of the position function).

Then, after you are given the velocity function (the derivative, s'(t)), you can find the velocity at t=5, by simply plugging t=5 into s'(t).