Velocity and acceleration.

  • Thread starter fahd
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velocity & acceleration PLZZZZ HELP ME!

:eek: hi i have this question from fowles anlaytical mechanics.It says

A racing car moves along a circular track of radius 'a'.The speed of the car varies with time as v=kt where k is a poistive constant.Show that the angle between the velocity vector and acceleration vector is 45 degrees when t= (a/k)^1/2

i took the equation of motion to be r= a cos[(kt^2)/a]+a sin[(kt^2)/a]
then differentiated to get velocity and acceleration...i represented a particular portion of each velocity and acceleration as a unit vector p and q so as to reduce the size of the equations.HOwever when i try to calculate the angle between a and v..at that time..it doesnt come out to be 45..Can ne one please help ,me!!
 
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Tide
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You can determine both the tangential and radial (centripetal) components of velocity and acceleration at any instant. Make it easy on yourself and choose your coordinates for a given instant of time to be such that the tangential direction is, say, in the postitive y direction and the radial component in the x direction.

Then note that [itex]\vec v \cdot \vec a = v a \cos \theta[/itex] and you can explicitly evaluate the dot product since you know the components of each vector. What must t be in order for [itex]\theta[/itex] to be 45 degrees?
 

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