Calculating velocity at a certain time

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SUMMARY

The velocity of a body described by the position function X(t)=9·cos(1.2t)·e^{-t/10} at t=17 seconds is calculated using the derivative formula v(t)=dr/dt. The resulting velocity is -1.975, indicating that the body is moving in the opposite direction. This negative value is consistent with the vector nature of velocity, which encompasses both magnitude and direction. The calculation assumes that the frequency 1.2 in the cosine function is in radians per second.

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Homework Statement


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

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

2. The attempt at a solution
I'm trying to use the formula [itex]v(t)=\frac{dr}{dt}[/itex] in order to get the velocity and I get: [itex]-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[/itex]

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: [itex]X(t)=9\cdot cos(1.2t)\cdot e^{-t/10}[/itex].

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

2. The attempt at a solution
I'm trying to use the formula [itex]v(t)=\frac{dr}{dt}[/itex] in order to get the velocity and I get: [itex]-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[/itex]
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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