Question #5def0

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Question #5def0

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Answer 1 · 2016-08-25T11:46:43

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

Explanation:

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

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

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

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

The path lenght is given by

$d = \int_{0}^{t_{f}} d s = \int_{0}^{2 π} \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2}} d t$ $d = int_0^(t_f)ds = int_0^(2pi)sqrt((dx/dt)^2+(dy/dt)^2)dt$ so

$d = \int_{0}^{t_{f}} (\sqrt{t^{2} + 2 t + 2}) d t$ $d = int_0^(t_f)(sqrt(t^2+2t+2))dt$ which gives

$d = 26.9651$ $d=26.9651$

but $v = \frac{d}{t_{f}} = 4.29163 [m/s]$ $v = d/(t_f)=4.29163["m/s"]$

Attached a figure showing the point $A$ $A$ path.

enter image source here

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