Car Traveling the Circle - Kinematic Quiz

AI Thread Summary
The discussion revolves around a physics problem involving a car traveling in a circular path at a constant speed of 10 m/s, with specific angles at points A and B. The original poster struggles with understanding how acceleration can exist despite constant speed, as well as how to calculate the components of velocity and acceleration. Responses clarify that constant speed does not equate to zero acceleration due to the change in direction of velocity in circular motion. The conversation also highlights the importance of grasping vector concepts and encourages self-learning rather than seeking direct answers. Overall, the thread emphasizes the complexities of kinematics in circular motion and the necessity of understanding underlying principles.
jtigers1
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Good evening.
I'd like to first start by saying I am new to this forum, and I am a physics noob. I know that very fact will help people shy away from trying to help me answer this question. This problem (which an image is posted) is 1 of 5 questions on a take home kinematic quiz. I've managed to answer all of them (correctly, I hope) except for this one.

One of the issues, I am having with this question is that I've never seen a similar problem. Also, some of the language in the problem may be confusing me.

Homework Statement



A car is traveling the circle shown between points A + B at a constant speed of 10 m/s. The radius of the circle is 100m.

There is a theta sub a (30 degrees) and a theta sub b (60 degrees).

I wrote everything that I could, but I've attached an image that shows everything.

a) Find the components of V_a and V_b

b) Compute delta V (keep in component form)

c) If it takes 10s to go between points A and B, what is the acceleration, in component form and magnitude angle form?

Relevant equations
a= delta v/delta t
components of vectors

The Attempt at a Solution



First off, the problems states the car is traveling at a constant speed, so doesn't that mean acceleration = 0?

But part c wants me to find the acceleration...? If its going at a constant speed, how is there a change in velocity for part b? These two things are confusing the crap out of me.

So far, I've started to find the components of V_a (hopefully the right way)
V_ax = (cos30)(10)
V_ay = (sin30)(10)

If this is the correct way to figure out the components, then I think I can get parts A and B of this question. Any guidance would be greatly appreciated.
 

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This is a circular motion problem. While the car is traveling at constant speed along the circumference of the circle, its velocity and acceleration with respect to coordinate axes whose origin is at the center of the circle will vary with the position of the car between points A and B.

Have you studied circular motion in your physics course yet?
 
nope, we have not gone over circular motion. >.<
 
Traveling at a constant speed doesn't mean that the acceleration is zero. If the velocity vector changes direction, this also constitutes acceleration. Acceleration is defined as the rate of change of velocity with respect to time, and velocity is a vector whose direction can change. So when you are determining acceleration, you need to include the effects of both the speed and direction of the velocity vector.
 
jtigers1 said:
nope, we have not gone over circular motion. >.<

Knowledge of circulation motion is not necessarily required to solve the problem, if you have studied Vectors!
 
The replies from you guys are ridiculous. This in no way, shape or form would be able to help any physics noob.

After a couple of hours of reevaluating my approach trying to follow the "advice" from the forum, I asked my Asian friend to just solve it for me.

Thanks for the useless help. If you all would like to call it help. I have learned to keep an Asian friend in my back pocket.
 
"If this is the correct way to figure out the components, then I think I can get parts A and B of this question. Any guidance would be greatly appreciated. "

"First off, the problems states the car is traveling at a constant speed, so doesn't that mean acceleration = 0? "

"Traveling at a constant speed doesn't mean that the acceleration is zero."


"Thanks for the useless help."

Need I say more?
 
Well, here's hoping nothing happens to your Asian friend, because then you'll be stuck. BTW, make sure your friend is comfortable on your lap when he takes the exam for you.
 
SteamKing said:
Well, here's hoping nothing happens to your Asian friend, because then you'll be stuck. BTW, make sure your friend is comfortable on your lap when he takes the exam for you.

Asian...lap...suddenly, I have an irresistible image of a beautiful Siamese on his lap. That has to be one smart cat. :biggrin:

Seriously, OP, if you want to be spoonfed, go to Yahoo Answers and offer 10 points to the first sucker who'll help you with a full solution. But you'd better be ready to do it for all your problems, because you'll never learn this way. And what *are* you going to do when it comes time for your exams?
 
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