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x(t)=asin(ωt):y(t)=acos(ωt);z(t)=bt i have to find the velocity, the acceleration and then their absolute values. I did not understand how I can calculate the absolute value.

Mary
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  • You work here with 3-dimensional vectors. So synonyms for absolute value are magnitude or modulus. – Roman Hric Oct 14 '19 at 13:03
  • I have found Vx(t)=awcos(wt), Vy(t)=-awsin(wt) and Vz(t)=b, so the absolute value of the velocity? how can I calculate since the parameter is in the angle? – Mary Oct 14 '19 at 14:04

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By definition of velocity: $$\nu = (a\omega\cos(\omega t), -a\omega\sin(\omega t), b) \Longrightarrow |\nu| = \sqrt{a^2\omega^2+b^2}$$ By definition of acceleration: $$\alpha = (-a\omega^2\sin(\omega t), -a\omega^2\cos(\omega t), 0) \Longrightarrow |\alpha| = |a|\omega^2$$

Ali Ashja'
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  • oh thank you! if I have to find the absolute value of the velocity=(4t,-6t,8t) it is square root of (4^2+(-6)^2+8^2) ? or not? and then for the acceleration, the absolute value will be 0? – Mary Oct 14 '19 at 14:19
  • Yes, since it goes on line produced by $(2,-3,4)$, at constant velocity $\sqrt{116}$, obviously its acceleration is $0$ for sure. – Ali Ashja' Oct 14 '19 at 14:23
  • so the absolute value of velocity will be root of 116 and the absolute value of acceleration will be just 0? – Mary Oct 14 '19 at 14:24
  • Yes, that is what i was said. – Ali Ashja' Oct 14 '19 at 14:27
  • okay, thank you so much – Mary Oct 14 '19 at 14:28