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A bicyclist is riding to the left with a velocity of 14 m/s . After a steady gust of wind that lasts 3.5 s, the bicyclist is moving to the left with a velocity of 21 m/s

Assuming the acceleration is constant, what is the acceleration of the bicyclist?

User Greg Owen
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2 Answers

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He accelerated from 14 m/s to 21 m/s in 3.5s so


a= (v_1 - v_0)/(t_1 - t_0) =(21-14)/(3.5) = 2 \textrm{ m/s}^2

Answer: 2 meters per second per second

User FBidu
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7 votes

Answer:

2 m/s²

Explanation:

A bicyclist is riding to the left with a velocity of 14 m/s.

Initial velocity, u = 14 m/s

After a steady gust of wind that lasts, t = 3.5 s

The bicyclist is moving to the left with a velocity of 21 m/s

Final velocity, v = 21 m/s


a=(v-u)/(t)


a=(21-14)/(3.5)


a=2\tex{ m/s}^2

Hence, The acceleration of the bicyclist is 2 m/s²

User Edson
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