Velocity and acceleration problem

[gtfitzpatrick]

Homework Statement

the position vecter r of a particle at a time t is

r = icos t + jsin t +kt show that the magnitudes of the velocity and acceleration vectors are constant. Describe the motion

The Attempt at a Solution

velocity = i(-sint) + j(cost) +k

magnitude is $$\sqrt{(-sint)^2 + (cost)^2 + 1}$$ = $$\sqrt{2}$$

acceleration = i(-cos t) - j(sin t)

magnitude is $$\sqrt{(-sint)^2 + (-cost)^2 }$$ = $$\sqrt{1}$$ = 1

but I am not sure how to discribe the motion, at first i was thinking of an object moving at a constant velocity of $$\sqrt{2}$$ and to move at a constant velocity it acceleration isn't changing so should that not mean the acceleration is 0 instead of 1?

 

[vela]

You've shown that the particle has a constant speed, not a constant velocity. While the particle moves with constant speed, the direction it's moving continuously changes.

Ignore the z-component for right now. If x=cos t and y=sin t, what path would the particle follow in the xy plane?

 

[gtfitzpatrick]

im trying to think about it, it follows a wave, up and down? but can it follow a sin and cos wave at the same time? do they not start from different points? I'm a bit confused

 

[vela]

If you're given a value of t, you can calculate what x and y equal and plot that point on the xy-plane. Try doing that for, say, t=0, pi/6, pi/4, pi/3, pi/2, 2pi/3, 3pi/4, 5pi/6, pi.

 

[gtfitzpatrick]

thanks a million for the replies.

After plotting it out, i think its moving at constant speed,not accelerating around the axis in a circular motion of radius 1?

 

[vela]

It's moving with constant speed, but it's accelerating because the direction of the velocity is changing with time.

So now add the motion in the z direction back in. So while the x and y coordinates trace out a circle, the z coordinate increases in time at a constant rate. How would you describe that path?