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.

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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.

Mathematics
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\[\large v(t) = \dfrac{\mathrm d }{\mathrm dt}~s(t)\]Right?
You just need to evaluate \(v(2)\)
If I evaluate the formula at t=-2 I just get -14

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How did you get -14? Can you find derivative of -2 - 6t ?
I don't think I know how to find derivatives properly.
When it says derivative, I think derived from t equaling 2 and the formula -2 -6t. -2-6*(2)=-14
I don't think that's how it works though.
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 \(\dfrac{\mathrm d }{\mathrm dx}c = 0\)
So here, you have \(\dfrac{\mathrm d }{\mathrm dt}(-2-6t) ~~~=~~~ \dfrac{\mathrm d }{\mathrm dt}(-2) + \dfrac{\mathrm d }{\mathrm dt}(-6t) = ~?\)
(d/-2d)-(d/-2d)=0
?
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)?
I don't have a teacher
Can you tell me what you got, then we can work it out.
So you are self-taught?
Probably online school :) ^^
yes
If someone could offer me the answer to this question, I would be glad to discuss how it was found

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