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Physics racing velocity problem

  1. Sep 28, 2003 #1
    A racing car moves on a circle of constant radius b.If the speed of the carvaries with time t according to the equation v=ct where c is a positive constant, show that the angle between the velocity vector and the acceleration vector is 45 degree at time t=(b/c)^1/2
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  3. Sep 28, 2003 #2


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    Show us what you DO understand of this problem and what you have done (or tried to do)so far.
  4. Sep 28, 2003 #3
    v = ct , so a= c
    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
  5. Sep 28, 2003 #4


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    You are forgetting that the acceleration has 2 components. A centripetal, which is always 90 degrees angle with respect to the velocity, and the one you mentioned, which is in the same direction as the velocity.
  6. Sep 29, 2003 #5


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    This is a heck of a good problem!

    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)t2. Put that into the velocity vector and differentiate again to find the acceleration vector. Take the dot product of the two vectors and the lengths of each and then use
    "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)1/2.)

    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)1/2). The along the tangent has length equal to the derivative of speed.
    Of course, the velocity vector is always tangent to the circle.
  7. Jan 22, 2004 #6
    When I solve for u I get u = (1/2)(c/b)t2 - where did your 1/2 from integration go?
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