A) The Singapore Flyer is the world's largest Ferris wheel. Its diameter is 150 m and it rotates once every 30 min. Find the magnitudes of the average velocity at the wheel's rim, over a 6.0 min interval.

this is what I tried but my solution is wrong.
Θ = (6 min/30 min)x360degrees= 72 degrees.
the vector r_1 = 150i and vector r_2 = (150*cos(72)i + 150*sin(72)j)m = (46.53i + 142.7j)m
Δr = r_2 - r_1 = (-103.6i + 142.7j)m
|Δr| = √(-103.6)² + (142.7)² = 176.3 m
6 min = 360 s
|V| = |Δr|/Δt = 176.3 m/360 s = 0.49 m/s
So my answer is |V| = 0.5 m/s but my answer is incorrect. please help

Question

A) The Singapore Flyer is the world's largest Ferris wheel. Its diameter is 150 m  and it rotates once every 30 min. Find the magnitudes of the average velocity at the wheel's rim, over a 6.0 min interval.

this is what I tried but my solution is wrong.
Θ = (6 min/30 min)x360degrees= 72 degrees. 
the vector r_1 = 150i and vector r_2 = (150*cos(72)i + 150*sin(72)j)m = (46.53i + 142.7j)m 
Δr = r_2 - r_1 = (-103.6i + 142.7j)m
|Δr| = √(-103.6)² + (142.7)² = 176.3 m
6 min = 360 s
|V| = |Δr|/Δt = 176.3 m/360 s = 0.49 m/s
So my answer is |V| = 0.5 m/s but my answer is incorrect. please help

anonymous · Answer

|dw:1361012594256:dw|
r1 and r2 are radius vectors. The only correction you have to make is changing 150 to 75 (in the second step).

anonymous · Answer

ok. thank you so much.