A particle is moving clockwise in a circle of radius 1.54 m. At a certain instant, the magnitude of its acceleration is a = 26.0 m/s2,and the acceleration vector has an angle of θ = 50° with the position vector, as shown in the figure. At this instant, find the speed, v of this particle.

Answer 1

a*sin50=v^2/r solve for v

Answer 2

not the right answer

Answer 3

It seems to me that a*cos50 is the centripetal component of the acceleration. The position vector is in the radial direction.

a*cos50 = v^2/r = 16.71 m/s^2
v^2 = 25.74 m^2/s^2
v = 5.07 m/s

Answer 4

thanks