The bakery of a small Indiana town has just made a huge strawberry shortcake for the opening of the summer festival. The cake has a radius of 0.5 m. A ladybug lands on the cake and walks around the edge for a distance of 80 cm. What is the angular displacement of the ladybug?

1.6 radians

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16 degrees

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0.16 radians

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1.6 degrees

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Solution

The correct option is A 1.6 radians
The distance travelled by the ladybug, s = 80 cm = 0.8 m. The radius, r = 0.5 m.
The angular displacement is $θ = \frac{s}{r} = 0.8 / 0.5$ or $θ = 1.6$ radians

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The bakery of a small Indiana town has just made a huge strawberry shortcake for the opening of the summer festival. The cake has a radius of 0.5 m. A ladybug lands on the cake and walks around the edge for a distance of 80 cm. What is the angular displacement of the ladybug?

The correct option is A 1.6 radiansThe distance travelled by the ladybug, s = 80 cm = 0.8 m. The radius, r = 0.5 m.The angular displacement is θ=sr=0.8/0.5 or θ=1.6 radians

The correct option is A 1.6 radians
The distance travelled by the ladybug, s = 80 cm = 0.8 m. The radius, r = 0.5 m.
The angular displacement is $θ = \frac{s}{r} = 0.8 / 0.5$ or $θ = 1.6$ radians