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## Homework Statement

A point particle has a position vector r⃗ (t) as a function of time t, given by

r(t)=(1−t2)x^−2t(t+5)y^+8(t+2)z^.

where distances are in meters, and time t is in seconds. Now, let t=t1= 23 s.

What is the (smaller) angle between the velocity vector at time t1 and the z^ axis? (in degrees)

## Homework Equations

## The Attempt at a Solution

I took the derivative of the position vector to get -2tx-(4t+10)y+8z. At t=23 the y component of the velocity is -102, but I don't know what to do from there.

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