Question

A car is traveling around a horizontal circular track with radius r = 270 m at a constant speed v = 17 m/s as shown. The angle θA = 23° above the x axis, and the angle θB = 58° below the x axis.

1. What is the x component of the car’s acceleration when it is at point A?

2.What is the y component of the car’s acceleration when it is at point A?

3.What is the x component of the car’s acceleration when it is at point B?

4. What is the y component of the car’s acceleration when it is at point B?

Answer 1

I found the magnitude of the car's acceleration to be 1.07m/s^2. But all the tangential and radial angles are really confusing me at the moment =(. If anyone could help me or just briefly explain that would be great.

I've attached a picture of the circle.

Answer 2

Since the car is traveling at constant speed the acceleration will point toward the center of the circle. Since you know the magnitude and direction of the acceleration you can use trig to find the x and y components.

Answer 3

Would that be to find the components of velocity or acceleration? I'm guessing acceleration.

Answer 4

Because it's asking for both acceleration and velocity so =/

Answer 5

Yes, acceleration points towards the center of the circle. The velocity is tangential to the circle, and perpendicular to the acceleration

Answer 6

Could anyone help me a bit more in depth, I've been trying a whole bunch of combinations of trig but I just dont understand how the acceleration of the car relates to point A and B.

Answer 7

This drawing shows the acceleration at point A. The y component of the acceleration will be \[a_y=a*\sin \theta\]The x component will be \[a_x=a*\cos \theta\]

This drawing shows the velocity at point A

This vector is perpendicular to your acceleration vector, so \[\phi=90-\theta\]Then the x component of the velocity is\[v_x=v*\cos\phi\]and the y component is\[v_y=v*\sin \phi\]
Hope this helps. You'll figure out point B the same way

Answer 8

That helps a ton thanks =)