Question

position of an object at time t is given by s(t) = -2 - 6t. Find the instantaneous velocity at t = 2 by finding the derivative.

Answer 1

\[\large v(t) = \dfrac{\mathrm d }{\mathrm dt}~s(t)\]Right?

Answer 2

You just need to evaluate \(v(2)\)

Answer 3

If I evaluate the formula at t=-2 I just get -14

Answer 4

*2

Answer 5

How did you get -14?

Can you find derivative of -2 - 6t ?

Answer 6

I don't think I know how to find derivatives properly.

Answer 7

When it says derivative, I think derived from t equaling 2 and the formula -2 -6t. -2-6*(2)=-14

Answer 8

I don't think that's how it works though.

Answer 9

You can just take derivative of each term individually. For each term, use power rule: \(\dfrac{\mathrm d }{\mathrm dx}x^n = nx^{n-1}\)

For constant, you have \(\dfrac{\mathrm d }{\mathrm dx}c = 0\)

Answer 10

So here, you have \(\dfrac{\mathrm d }{\mathrm dt}(-2-6t) ~~~=~~~ \dfrac{\mathrm d }{\mathrm dt}(-2) + \dfrac{\mathrm d }{\mathrm dt}(-6t) = ~?\)

Answer 11

(d/-2d)-(d/-2d)=0

Answer 12

?

Answer 13

Seem you are not really familiar with basic derivative rules. Can you tell me what you learned recently (what teacher tried to teach you, etc)?

Answer 14

I don't have a teacher

Answer 15

Can you tell me what you got, then we can work it out.

Answer 16

So you are self-taught?

Answer 17

Probably online school :) ^^

Answer 18

yes

Answer 19

If someone could offer me the answer to this question, I would be glad to discuss how it was found