A bike riding at 22.4 m/s skids to come to a halt in 2.55 s. Conclude the skidding distance of the bike.

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Muskan Anand

7 months ago

Initial velocity of the bike, u=22.4 m/su=22.4 m/s Final velocity of the bike, v=0 m/sv=0 m/s Let the acceleration be aa of the bike.  Time taken, t=2.55 sect=2.55 sec Therefore, a=v−ut=0−22.42.55a=v−ut=0−22.42.55 Or a=−8.7 m/s2a=−8.7 m/s2 Now using the equation s=ut+12at2s=ut+12at2 Skidding distance, s=22.4×2.55+12(−8.7)×(2.55)2s=22.4×2.55+12(−8.7)×(2.55)2 =57.12−28.2=57.12−28.2 =28.92 m=28.92 m Thus, the skidding distance of the bike is 28.92 m28.92 m.

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