Question

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?

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