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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...?
 2 years ago
 2 years 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...?
 2 years ago
 2 years ago

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

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

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

Dido525 Group TitleBest ResponseYou've already chosen the best response.3
Oops cos(t)
 2 years 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?
 2 years ago

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

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

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

Study23 Group TitleBest ResponseYou've already chosen the best response.0
@zepdrix yup
 2 years 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 .
 2 years 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?
 2 years ago

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

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

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

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