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anonymous
 one year ago
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.
anonymous
 one year ago
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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geerky42
 one year ago
Best ResponseYou've already chosen the best response.0\[\large v(t) = \dfrac{\mathrm d }{\mathrm dt}~s(t)\]Right?

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0You just need to evaluate \(v(2)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I evaluate the formula at t=2 I just get 14

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0How did you get 14? Can you find derivative of 2  6t ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't think I know how to find derivatives properly.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When it says derivative, I think derived from t equaling 2 and the formula 2 6t. 26*(2)=14

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't think that's how it works though.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0You can just take derivative of each term individually. For each term, use power rule: \(\dfrac{\mathrm d }{\mathrm dx}x^n = nx^{n1}\) For constant, you have \(\dfrac{\mathrm d }{\mathrm dx}c = 0\)

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0So here, you have \(\dfrac{\mathrm d }{\mathrm dt}(26t) ~~~=~~~ \dfrac{\mathrm d }{\mathrm dt}(2) + \dfrac{\mathrm d }{\mathrm dt}(6t) = ~?\)

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0Seem you are not really familiar with basic derivative rules. Can you tell me what you learned recently (what teacher tried to teach you, etc)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't have a teacher

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you tell me what you got, then we can work it out.

geerky42
 one year ago
Best ResponseYou've already chosen the best response.0So you are selftaught?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Probably online school :) ^^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If someone could offer me the answer to this question, I would be glad to discuss how it was found
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