A tool is programmed to move, such that x=2cos3t and y=cos2t. Find the velocity of the tool when t=4.1s.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The velocity of the machine can be divided into two components, x and y.

It moves at x = 2 cos 3t in the x-direction.

Therefore the x component of the velocity is dx/dt = -2*3 sin 3t.

At t = 4.1, this is equal to 1.5793.

In the y- direction it moves at y = cos 2t.

dy/dt = -2* sin 2t

At t = 4.1, this is equal to -1.8814

The direction of the velocity is at angle of -0.8724 radians to the x-axis.

Therefore the magnitude of the velocity at t = 4.1 is sqrt ( 1.5793^2 + 1.8814^2) = 2.4563.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

We assume X and Yare perpendicular direction representing x and y axis.

The  component of the velocity of  the machine in x direction is at time t isgiven by: dx/dt = d/dt(2cos2t) = 2(-sin3t)3= -6sin2t.

The component of velocity of the velocity of the machine in y direction at time t is given by: dy/dt = d/dt(cos2t) = (-sin2t)(2) =  -2sin2t.

Therefore the v velocity  of the machine at time t is given by : v(t) =  {(x component  of the velocity)^2+(y component of the velocity)^2}^(1/2) =  (-6sin3t)^2 +(-2sin2t)^2}^(1/2) = {36(csin3t)^2+4sin2t)^2 }^(1/2)

v(t) = 2{9(sin3t)^2 + (sin2t)^2}^(1/2)

v(4.1) = 2{9sin12.3))^2 + (sin8.2)^2}^(1/2)

v(4.1) = 2{0.62362+0.88497}^(1/2)

v(t) =  2.4565 length units/sec is the velocity of the machine tool.

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