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anonymous
 4 years ago
The equation of motion is \(\ s=2costy+3sint, t\le0 \). Find velocity/acceleration at time t. I took the derivative to find the velocity, but got 0...?
anonymous
 4 years ago
The equation of motion is \(\ s=2costy+3sint, t\le0 \). Find velocity/acceleration at time t. I took the derivative to find the velocity, but got 0...?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Good catch! That's supposed to be a t only.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so \[s=2\cos(y) + 3\sin(t) , t \le0\] ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well the velocity at time t is just the derivative. Can you differentiate that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Dido525 I tried, but I got 0, which doesn't seem correct.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lets see:dw:1351660147750:dw

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.0Oh that was a 2? :P sorry..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so Plug in your value for t and you have your answer :P .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Would I add the derivates of each here? What happens to the constants?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just did the derivative for you O_o .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I suppose I was confused by the constants...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ohh. If you have constants out in from you take out the constant, take the derivative of the function and multiply the constant back in.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ohh.Okay. I was getting the product rule confused with the addition one. So for the acceleration, I take the derivative of the derivative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay!! Thanks so much!!!
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