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a bicyclist in the Tour De France crests a mountain pass as he moves at 15km/h. At the bottom 4km/h farther, his speed is 75km/h. What is his average acceleration(m/s^2) while riding down the mountain?
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this is how i worked it out but it it didn't work: Vi=15km/h-->4.1m/s V=75km/h--> 20.8m/s and s=4km--> 4000m so 2a=(V^2-Vi^2)/s 2a=(20.8-4.1)/4000
it is 20.8^2 & 4.1^2
what do you mean by "At the bottom 4km/h farther" ? Is it distance? we don't measure distance in km/h
@rajathsbhat thx, and @ujjwal that was a typo..
0.052m/s^2 ?
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okay, found the answer :)
congrats! but how did you find it?
umm, well raj corrected by square mistake so:2a= (20.8^2-4.1^2)/4000 a=0.104/2=0.52m/s^2
@bronzegoddess a = 0.104/2 = 0.052 ms^-2
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