A bicyclist in the Tour de France crests a mountain pass as he moves at 18km/h. At the bottom, 3.6km farther, his speed is 60km/h. What was his average acceleration (in m/s^2) while riding down the mountain?

Question

Alexa Freeman

Physics

17 January, 18:18

A bicyclist in the Tour de France crests a mountain pass as he moves at 18km/h. At the bottom, 3.6km farther, his speed is 60km/h. What was his average acceleration (in m/s^2) while riding down the mountain?

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Answers (1)

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Wyatt Rose · Answer 1

For this case we can use the following kinematic equation:

Vf ^ 2 = Vo ^ 2 + 2 * a * d

Where,

Vf: final speed.

Vo: initial speed.

a: acceleration.

d: distance.

We cleared the acceleration:

Vf ^ 2 = Vo ^ 2 + 2 * a * d

Vf ^ 2-Vo ^ 2 = 2 * a * d

a = (Vf ^ 2-Vo ^ 2) / (2 * d)

Substituting the values:

a = ((60) ^ 2 - (18) ^ 2) / (2 * (3.6))

a = 455 Km/h^2

Making change of units:

a = 455 * (1000 / (3600) ^ 2)

a = 0.035108025 m / s ^ 2

Answer:

his average acceleration (in m/s^2) while riding down the mountain is:

a = 0.035108025 m / s ^ 2