Question

A racing car moves on a circle of constant radius b. the speed of th car varies with time t according to v=ct where c is a positive constant. find the time t when the angle between the velocity vector and the acceleration vector is 45 degrees

Answer 1

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.)