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Melawrghk
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Homework Statement
The shape of the arch is defined by a sine function with a = 0.2 and the width of the arch is 1m. The velocity v is constant at 2m/s. What is the velocity at point P when x=0.25m?
The Attempt at a Solution
So since v is just in the x-direction, I figured that's the vx speed.
Then I came to having to find an equation to vy and this is where I get stuck. I remember from high school physics that equation of velocity can be essentially found by taking a derivative of the position equation.
In this case I get:
vy=y'=0.2*pi*cos(pi*x)
From which I can find that at x=0.25m, vy = 0.444288 m/s
But that combined with the horizontal speed of 2m/s gives me a resultant velocity of 2.05ish, instead of 2.19 that I'm supposed to get. Help please?
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