Circular motion x, y acceleration

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aruji73
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Can you tell me if my answers and sign are correct ?

A car is traveling around a horizontal circular track with radius r = 250.0 m at a constant speed v = 18.0 m/s as shown. The angle θA = 19.0° above the x axis, and the angle θB = 60.0° below the x axis.

What is the x component of the car’s acceleration when it is at point B = 0.527

What is the y component of the car’s acceleration when it is at point B = 1.1859

As the car passes point B, the y component of its acceleration is-increasing
 

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Welcome to PF;
Well you left off the units ... that's marks lost right there.
How did you come by those numbers?

Note: it is possible to get the right answer despite misunderstanding the problem... but if you understand the problem, you'll probably also be confident with your answer. s there some special reason you are concerned about your answers?
 
I am concerned if my signs are right
 
Can you tell me if my answers and sign are correct ?

A car is traveling around a horizontal circular track with radius r = 250.0 m at a constant speed v = 18.0 m/s as shown. The angle θA = 19.0° above the x axis, and the angle θB = 60.0° below the x axis.

What is the x component of the car’s acceleration when it is at point B = 1.296*cos60
What is the y component of the car’s acceleration when it is at point B = 1.296*sin60

are both the magnitudes going to have a positive value ?
 
Which direction does the acceleration vector point in circular motion at constant speed?
Draw the acceleration vector arrow on your diagram.

When you have done that, you will know if your signs are correct.

You have still not provided units.
You have still not said how you got those numbers: where did 1.296 come from?
I'm not doing this just to be pedantic or make your life difficult! What I am trying to do is direct your attention to where the answer to your question lies. The idea is that, next time, you won't have to ask.