Given functions, determine which has a velocity of zero at t=0 s| University Physics

The expressions below give the position function for an object moving along the x-axis. In which case is the velocity zero at the time t=0\:s. In all cases, A,\:B,\:\omega are constants.

  1. x\left(t\right)=A\:cos\left(\omega t+\frac{\pi }{2}\right)

  2. x\left(t\right)=A\:sin\left(\omega t+\frac{\pi }{2}\right)

  3. x\left(t\right)=A\:e^{-\omega t}

  4. x\left(t\right)=At^2+Bt

  5. x\left(t\right)=\frac{A}{\left(t+\omega ^{-1}\right)}

ANSWER: x\left(t\right)=A\:sin\left(\omega t+\frac{\pi }{2}\right)

To answer this question, we need to get the velocity function by getting the derivative of the given position functions. After getting the derivative, the velocity function with a value of zero at time t=0 is the answer. 

If we take the function x\left(t\right)=A\:sin\left(\omega t+\frac{\pi }{2}\right), and find its derivative, the resulting function is 

v\left(t\right)=A\:\omega \:cos\left(\omega t+\frac{\pi }{2}\right).

If we substitute t=0 in the velocity function, the result is 

v\left(t\right)=A\:\omega \:cos\left(\frac{\pi }{2}\right).

And we know that the cos\left(\frac{\pi }{2}\right) is equal to zero, which makes the velocity function equal to zero at time zero. 

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