An object traveling a circular path of radius 5 m at constant speed experiences an acceleration of 3 m/s2. if the radius of its path
is increased to 10 m, but its speed remains the same, what is its acceleration?
0.3 m/s2
1.5 m/s2
6 m/s
12 m/s2

answer:

explanation:

acceleration = (velocity^2)/radius. if the radius doubles, then as a result the acceleration halves. 

3^2 m/s^2 = velocity^2 m/s / 5 m -> 45 m^2/s^2 = velocity^2 m/s -> velocity = 6.71 m/s

acceleration = (6.71)^2 m/s / 10m = 4.5m/s^2

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west