# A boat tied up at a dock bobs up & down with the passing waves. Vertical distance between its high & low points is 1.8 m.The cycle is repeated every 4 s. At what time or times during...

A boat tied up at a dock bobs up & down with the passing waves. Vertical distance between its high & low points is 1.8 m.

The cycle is repeated every 4 s. At what time or times during a cycle is the instantaneous rate of change of the vertical position of the boat equal to 0 and at what times is it a maximum?

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### 1 Answer

The boat moves up and down with the passing waves. I take the height of the boat to be a sinusoidal function. And the height of the boat is zero at the start of the cycle.

The distance between the high and the low points is 1.8. So the amplitude of the vertical displacement is 1.8/2 = 0.9 m.

As the cycle is repeated every 4 s, the time period is 4s.

The equation of the motion of the boat as it bobs up and down is given by f(t) = 0.9*sin ((t - 4)*0.5*pi)

The rate of change of the vertical velocity is the derivative of f(t), f'(t) = 0.9*0.5*pi*cos ((t - 4)*0.5*pi).

The instantaneous rate of change of velocity is 0 when 0.9*0.5*pi*cos ((t - 4)*0.5*pi) = 0

or at t = 1 + n*4 and at t = 3 + n*4

The instantaneous rate of change is maximum at t = 0 + 4*n and at t= 2 + n*4

Therefore the instantaneous rate of change in the vertical position is :

**0 at t = 1 + n*4 and t = 3 + n*4 and maximum at t = 0 + 4*n and t= 2 + n*4**