Physics racing velocity problem

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Homework Help Overview

The problem involves a racing car moving in a circular path with a radius b, where the speed of the car varies with time according to the equation v=ct. The objective is to demonstrate that the angle between the velocity vector and the acceleration vector is 45 degrees at a specific time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the components of acceleration, including centripetal and tangential components, and their relationship to velocity. There are attempts to express the velocity and acceleration vectors in terms of time and to analyze their geometric relationship.

Discussion Status

Multiple approaches to the problem have been presented, with some participants offering different methods for deriving the angle between the vectors. There is an ongoing exploration of the mathematical relationships involved, and some participants are questioning specific steps in the reasoning process.

Contextual Notes

Participants are navigating through the complexities of circular motion and the implications of varying speed, while also addressing potential misunderstandings about vector components and integration in their calculations.

newton1
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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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Show us what you DO understand of this problem and what you have done (or tried to do)so far.
 
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
 
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.
 
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.
 
Originally posted by HallsofIvy
The length of that vector is bu'(t) and that must equal ct: bu'= ct so
u(t)= (c/b)t2.

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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