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?
goutham 1995 gumm , 13 Years ago
Grade 12
2 Answers
Chetan Mandayam Nayakar
Last Activity: 13 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
Last Activity: 2 Years 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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