Question

A car initially traveling eastward turns north by traveling in a circular path at uniform speed as shown in the figure below. The length of the arc ABC is 230 m, and the car completes the turn in 41.0 s. (a) Determine the car's speed. 5.609m/s (b) What is the magnitude and direction of the acceleration when the car is at point B? magnitude m/s2 direction counterclockwise from the +x-axis

Answer 1

In order to do this, re-call that :

(a) velocity = displacement/time speed = distance/time

For this your distance covered is 230m ( the quarter of the circle ) and the time was 41 s.

avg_v = 230/41 = 5.609 m/s

(B)

The circumference of the circle is given by 230 * 4 = 920 m therefore its radius = 920/π/2 = 146.49 m


acceleration in a circular motion is defined by a = v^2 / r

You would just plug your values in a = 5.609 ^2 / 146.49 = 0.214 m/s²

acceleration = 0.214 m/s²

and the direction will be towards the center of the curve which will be perpendicular to the instantaneous velocity.
35° + 90° = 125°