A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h.

Question

A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. 4.0 km down the hill his speed is 75 km/h. What was his average acceleration (in m/s^2) while riding down the mountain? A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. 4.0 km down the hill his speed is 75 km/h. What was his average acceleration (in m/s^2) while riding down the mountain? A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. 4.0 km down the hill his speed is 75 km/h. What was his average acceleration (in m/s^2) while riding down the mountain?

Explanation / Answer

Initial speed of the bicyclist (u)= 18 km/h = 18*1000/3600 = 5 m/s
Final speed of the bicyclist(V) = 75 km/h = 75*1000/3600 = 20.833 m/s
Distance covered (S)= 4 km = 4000 m
So we know from the third equation of motion
V2 = u2 + 2aS
(20.833)2 = (5)2 + 2a*4000
therefore accelearation (a) = 662.625 km/hr2 = 0.051128 m/s2

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