Question

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?

Answer 1

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

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

Answer: 2 meters per second per second

Answer 2

Answer:

2 m/s²

Step-by-step 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

$a=\dfrac{v-u}{t}$

$a=\dfrac{21-14}{3.5}$

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

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