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Study23
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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...?
 one year ago
 one year ago
Study23 Group Title
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...?
 one year ago
 one year ago

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Dido525 Group TitleBest ResponseYou've already chosen the best response.3
is that 2cos(ty) ?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Good catch! That's supposed to be a t only.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.3
so \[s=2\cos(y) + 3\sin(t) , t \le0\] ?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.3
Oops cos(t)
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.3
Well the velocity at time t is just the derivative. Can you differentiate that?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
@Dido525 I tried, but I got 0, which doesn't seem correct.
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.3
Lets see:dw:1351660147750:dw
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Oh that was a 2? :P sorry..
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix yup
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.3
so Plug in your value for t and you have your answer :P .
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Would I add the derivates of each here? What happens to the constants?
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.3
I just did the derivative for you O_o .
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
I suppose I was confused by the constants...
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.3
Ohh. If you have constants out in from you take out the constant, take the derivative of the function and multiply the constant back in.
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Ohh.Okay. I was getting the product rule confused with the addition one. So for the acceleration, I take the derivative of the derivative?
 one year ago

Study23 Group TitleBest ResponseYou've already chosen the best response.0
Okay!! Thanks so much!!!
 one year ago

Dido525 Group TitleBest ResponseYou've already chosen the best response.3
Welcome :) .
 one year ago
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