Question

A car is traveling around a horizontal circular track with radius r = 240 m as shown. It takes the car t = 53 s to go around the track once. The angle θA = 19° above the x axis, and the angle θB = 61° below the x axis. What is the x component of the car’s velocity when it is at point A? What is the y component of the car’s velocity when it is at point A

Answer 1

Here is an image to illustrate the question: http://assets.openstudy.com/updates/attachments/4e90b26b0b8bfb9f23e2360e-mattyc33-1318105843410-caroncurve2.png

Ive been stuck on those two questions for 2 hours now,somebody please help

Answer 2

I have tried using trig to solve but either im not plugging in the right numbers in the correct way or I am suppose to do something else...any help will be greatly appreciated.

Answer 3

At any point, the velocity vector is tangential to the path traced by the object.

@joeymiller789
Try to find the angle between the velocity vector at point A and the x-axis.

Answer 4

\[v=2\pi*r \div t\]
\[v = 28.45m/s\]

so \[v_y=v \sin (71)\]
\[v_y=27m/s\]