The position of an object at time t is given by s(t) = -9 - 5t. Find the instantaneous velocity at t = 4 by finding the derivative

Question

ganeshie8 · Answer

yes so lets find the derivative

ganeshie8 · Answer

$\large  s(t) =  -9-5t$

$\large  v(t) =  s'(t) = ?$

anonymous · Answer

ummm how do I work this?

ganeshie8 · Answer

$$\large   \dfrac{d}{dx}(x^n) = nx^{n-1}$$

seen this formula before ?

anonymous · Answer

I think so

ganeshie8 · Answer

and you may be knowing that derivative of a constant is 0 :
$$\large \dfrac{d}{dx}(c) =  0$$

anonymous · Answer

Okay

ganeshie8 · Answer

$$\large  v(t) =  \dfrac{d}{dt}(s(t)) =   \dfrac{d}{dt}(-9-5t)$$

ganeshie8 · Answer

whats the derivative of "-9"  ?

anonymous · Answer

0?

ganeshie8 · Answer

yes!
what about the derivative of "-5t" ?

ganeshie8 · Answer

"-5t" looks like "y=mx" right ?
so derivative is just "-5"

ganeshie8 · Answer

$$\large  v(t) =  \dfrac{d}{dt}(s(t)) =   \dfrac{d}{dt}(-9-5t) $$

$$\large   =   0-5 $$
$$\large   =   -5 $$

ganeshie8 · Answer

$$\large v(t) =   -5 $$

ganeshie8 · Answer

see if that makes more or less sense..