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?
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
