A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 3.8 km farther, his speed is 85 km/h. What was his average acceleration (in m/s2) while riding down the mountain?

Question

Physics

Essence

8 August, 19:27

A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 3.8 km farther, his speed is 85 km/h. What was his average acceleration (in m/s2) while riding down the mountain?

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Answers (1)

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Kelton French · Answer 1

Initial speed, u = 18 km/h

Final speed, v = 85 km,/h

Distance traveled, s = 3.8 km

Note:

1 km/h = 0.27778 m/s

Therefore

u = 18*0.27778 = 5 m/s

v = 85*0.27778 = 23.61 m/s

s = 3.8 km = 3800 m

Let a = the acceleration.

Use the formula

v² = u² + 2as, or

a = (v² - u²) / 2s

= (557.493 - 25) / 7600

= 0.07 m/s²

Answer: 0.07 m/s²