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newton1

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- Thread starter newton1
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newton1

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- #2

HallsofIvy

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Show us what you DO understand of this problem and what you have done (or tried to do)so far.

- #3

newton1

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i think is a= a(t) + a (n)

a(t) = a sin(b) and a(n)= a cos(b)

and a(t) = a(n) when b is 45 degree

- #4

krab

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HallsofIvy

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I did it two different ways:

The hard way: Since the car is going around a circle of radius b, it "position vector"(relative to the center of the circle) is of the form b cos(u(t))i+ b sin(u(t))j where u(t) is some function of time.

The velocity vector is, therefore, -bu' cos(u(t))i+ bu' cos(u(t))j.

The length of that vector is bu'(t) and that must equal ct: bu'= ct so

u(t)= (c/b)t

"a*v= |a||v| cos(theta)" to find the angle between them. (a*v and |v| turn out to be easy. |a| is more complicated but reduces easily at t= (b/c)

The easy way: As krab said, look at the components of the acceleration vector, one parallel to the tangent to the circle, the other perpendicular to it.

The component perpendicular to the tangent (toward the center of the circle) is exactly the same as for a constant speed (use the speed AT t= (b/c)

Of course, the velocity vector is always tangent to the circle.

- #6

xsamx

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Originally posted by HallsofIvy

The length of that vector is bu'(t) and that must equal ct: bu'= ct so

u(t)= (c/b)t^{2}.

When I solve for u I get u = (1/2)(c/b)t

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