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when jim and rob rides bicycle, jim can only accelarate at three quarters the accelaration of rob. both start from rest at the bottom of a long straight roadwith a constant uppward slope. if rob takes 5 mins to reach the top, how much earlier should jim start to reach the top at the same time as rob?

when jim and rob rides bicycle, jim can only accelarate at three quarters the accelaration of rob. both start from rest at the bottom of a long straight roadwith a constant uppward slope. if rob takes 5 mins to reach the top, how much earlier should jim start to reach the top at the same time as rob?


 

Grade:12

2 Answers

Chetan Mandayam Nayakar
312 Points
12 years ago

let d be distance, acceleration of jim=a

d=(1/2)(4a/3)(5min)2

let t be time for jim

d=(1/2)(a)(t)2

 

Evanta Bafna
22 Points
one year ago
Accel of jim is “a” and of rob is “3a/4”
Time of jim is 5min and of rob is t
S covered is constant
S=1/2at^2
ie s directly proportional to at^2
S1/S2=(at2 of jim)/(at2 of rob)
s1/s2=1 therefore, on putting values t2 of rob comes out 5.77min
extra timing is 0.77 min ie 46.41s

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