Acceleration on a parabolic curve

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issacnewton
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Hi

I am doing some problem in Hibbeler's Engineering Dynamics (12 ed.). I have posted the problem as an attachment. I think the author has not given the x coordinate of the point B. Once that is given we can use the radius of curvature formula

[tex]\rho = \frac{[1+(dy / dx)^2]^{3/2}}{|d^2y/dx^2| }[/tex]

to get the radius of curvature at point B. And then we can find normal and tangential components of the acceleration at point B. But to use the above formula,we need to know the x coordinate of the point B. And I don't know how to find that from the given information.
Any ideas ?

thanks
 

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Hi ehild

Yes, I can find [itex]dy/dx[/itex] alright, but we have been given an arc length from point A to B. x coordinate of A is 100. So I tried to set up an equation using formula for arc length from calculus, and tried to integrate it to get some transcendental equation. Mathematica gave me very ugly output...

thanks
 
The speed is given as function of s. The speed is ds/dt. Knowing s, you can find the time when the car is at point B. Try to use it...I do not know the solution yet.

ehild
 
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If you are trying to find the distance traveled on the parabola, the formula for the radius of curvature is inappropriate. There is another was to find arclength from calculus, which is given in differential form:

ds^2 = dx^2 + dy^2

dividing thru by dx^2:

[ds/dx]^2 = 1 + [dy/dx]^2

taking square roots:

ds/dx = sqrt(1 + [dy/dx]^2)

ds = sqrt(1 + [dy/dx]^2) * dx

Integrate both sides and you will have your arclength.
 
The arclength is given. The x coordinate is the question. From the given s=51.5 m, x can be calculated numerically (it is ugly, I admit, but close to 50 m).

ehild
 
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