Physics racing velocity problem

In summary, a racing car moves on a circle of constant radius b and the speed of the car varies with time t according to the equation v=ct where c is a positive constant. The angle between the velocity vector and the acceleration vector is 45 degrees at time t=(b/c)^1/2. Two different approaches can be taken to solve this problem, one involving finding the position and velocity vectors and using dot product to find the angle, and the other involving looking at the components of the acceleration vector parallel and perpendicular to the tangent of the circle.
  • #1
newton1
152
0
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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  • #2
Show us what you DO understand of this problem and what you have done (or tried to do)so far.
 
  • #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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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?
 

1. What is the formula for calculating velocity in physics racing problems?

The formula for calculating velocity is v = d/t, where v represents velocity, d represents distance, and t represents time.

2. How do I convert velocity from miles per hour (mph) to meters per second (m/s)?

To convert from mph to m/s, you can use the conversion factor 0.44704. Multiply the velocity in mph by 0.44704 to get the equivalent velocity in m/s.

3. How does acceleration affect velocity in physics racing problems?

Acceleration is the rate of change of velocity. This means that if acceleration is positive, velocity will increase over time. Conversely, if acceleration is negative, velocity will decrease over time.

4. What is the difference between speed and velocity in physics racing problems?

Speed is the measure of how fast an object is moving, while velocity is the measure of how fast an object is moving in a specific direction. This means that velocity takes into account both the speed and direction of an object's motion.

5. How can I use physics principles to improve my racing performance?

Understanding the concepts of velocity, acceleration, and force can help you make more informed decisions while racing. By analyzing the physics of your movements and the track, you can adjust your techniques and strategies to optimize your performance.

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