Grade 12Algebra
answer..................>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
answer..................>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
2 Answers
Samyak Jain
The particle is moving in a 3-D coordinate system.
Displacement in x-direction is given as x = a cos
t.
Differentiate both sides wrt t to get component of velocity in x-direction.
Similarly, y = a sint 
y = dy/dt = acost
z = at z = dz/dt = a
Speed of the particle v is the magnitude of the vector sum of the velocity components.
x,y and z axes are mutually perpendicular.
= ![\sqrt{(a\omega)^2[sin^2 \omega t + cos^2\omega t + 1]}](https://latex.codecogs.com/gif.latex?%5Cdpi%7B80%7D%20%5Csqrt%7B%28a%5Comega%29%5E2%5Bsin%5E2%20%5Comega%20t%20+%20cos%5E2%5Comega%20t%20+%201%5D%7D)
= a
= 
[
sin2t + cos2t = 1]
Thus, the speed of the particle is .
Vikas TU
If we differentiate the x,y,z co-ordinates wrt time we will get the velocity to be
-awsinwt, awcoswt , aw. So if we try to find the magnitude of this we will get the answer to be
1.732*aw as the speed.
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