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Answer 1 · 2016-08-25T11:46:43

$v =4.29163["m/s"]$

The point $p={x,y}$ given by

$p = {(t+1)cos(alpha t + pi/2),(t+1)sin(alpha t+pi/2)}$ describes the point $A$ path.

Here we will assume $alpha = 1$ . This path, begins for $t = 0$ in

$p_i = {0,1}$ and after a time $t_f = 2pi$ , the point $A$ location is
$p_f = {0,2pi+1}$

The path lenght is given by

$d = int_0^(t_f)ds = int_0^(2pi)sqrt((dx/dt)^2+(dy/dt)^2)dt$ so

$d = int_0^(t_f)(sqrt(t^2+2t+2))dt$ which gives

$d=26.9651$

but $v = d/(t_f)=4.29163["m/s"]$

Attached a figure showing the point $A$ path.

enter image source here