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        answer..................>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
6 months ago

Samyak Jain
333 Points
							The particle is moving in a 3-D coordinate system.Displacement in x-direction is given as x = a cos$\dpi{100} \omega$t.Differentiate both sides wrt t to get component of velocity in x-direction.$\dpi{100} \Rightarrow$ $\dpi{80} v$x = dx/dt = a(–sin$\dpi{100} \omega$t) . $\dpi{100} \omega$ = – a$\dpi{100} \omega$sin$\dpi{100} \omega$t    [$\dpi{100} \because$ a is constant]Similarly, y = a sin$\dpi{100} \omega$t  $\dpi{100} \Rightarrow$ $\dpi{80} v$y = dy/dt = a$\dpi{100} \omega$cos$\dpi{100} \omega$tz = a$\dpi{100} \omega$t  $\dpi{100} \Rightarrow$  $\dpi{80} v$z = dz/dt = a$\dpi{100} \omega$Speed of the particle v is the magnitude of the vector sum of the velocity components.x,y and z axes are mutually perpendicular.$\dpi{80} \therefore$ v = $\dpi{80} \sqrt{v_{x}^2 + v_{y}^2 + v_{z}^2}$ = $\dpi{80} \sqrt{(-a\omega sin\omega t)^2 + (a\omega cos\omega t)^2 + (a\omega)^2}$                                 = $\dpi{80} \sqrt{(a\omega)^2[sin^2 \omega t + cos^2\omega t + 1]}$ = a$\dpi{100} \omega$$\dpi{80} \sqrt{1+1}$ = $\dpi{80} \sqrt{2}$ $\dpi{100} \omega$[$\dpi{80} \because$ sin2$\dpi{100} \omega$t + cos2$\dpi{100} \omega$t = 1]Thus, the speed of the particle is $\dpi{80} \sqrt{2}$ $\dpi{100} \omega$.

6 months ago
Vikas TU
8748 Points
							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.

6 months ago
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