Calculating velocity at a certain time

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The position of a body is described by the function X(t)=9·cos(1.2t)·e^{-t/10}. To find the velocity at t=17 seconds, the derivative v(t)=dr/dt was calculated, resulting in a value of approximately -1.975. The negative velocity indicates that the body is moving in the opposite direction, which is a normal aspect of vector quantities. The discussion also highlights the importance of interpreting the units of the angular frequency in the cosine function. Overall, the calculation and interpretation of negative velocity are confirmed as correct.
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Homework Statement


The following describes the position of a body as a function of time: X(t)=9\cdot cos(1.2t)\cdot e^{-t/10}.

What is the velocity of the body at t=17sec ?

2. The attempt at a solution
I'm trying to use the formula v(t)=\frac{dr}{dt} in order to get the velocity and I get: -9\cdot e^{\frac{-17}{10}}\cdot [sin(1.2\cdot 17)\cdot 1.2+\frac{1}{10}\cdot cos(1.2\cdot 17)]=-1.975

I guess negative velocity means the body is traveling in the oposite direction but is seems a little odd. Could this be the right answer?
 
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Hello phyyy,

Welcome to Physics Forums!

phyyy said:

Homework Statement


The following describes the position of a body as a function of time: X(t)=9\cdot cos(1.2t)\cdot e^{-t/10}.

What is the velocity of the body at t=17sec ?

2. The attempt at a solution
I'm trying to use the formula v(t)=\frac{dr}{dt} in order to get the velocity and I get: -9\cdot e^{\frac{-17}{10}}\cdot [sin(1.2\cdot 17)\cdot 1.2+\frac{1}{10}\cdot cos(1.2\cdot 17)]=-1.975
That's what I got (ignoring any rounding differences). :approve: But that assumes that the number 1.2 in the cosine function has units of rad/sec (as opposed to deg/sec).
I guess negative velocity means the body is traveling in the oposite direction but is seems a little odd. Could this be the right answer?
That shouldn't seem that odd. Velocity is vector. It has both magnitude and direction.
 
Thanks collinsmark!
 
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