from the same place at 7 a.m. A started walking in the north at the speed of 5 km/hr . After 1 hour B started cycling in the east at a speed of 16 km /hr . At what time will they be at a distance of 52km apart frim each other ?
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Let t=travel time of the bike
then
(t+1)=walking Time
dist=speed *time
This is a pythag problem :a^2+b^=c^2, where
a=16t, bike dist
b=5(t+1), walking dist
c=52, dist apart in t hrs from 8 AM
(16t)^2+(5(t+1) ^2 =52^2
256t^2+(5t+5) ^2 =2704
256t^2+25t^2+50t+25=2704
281t^2+50t+25-2704=0
281t^2+50t-2679=0
solve this equation using the quadratic formula
positive solution t=3hrs
At 11Am they will be 52km apart
then
(t+1)=walking Time
dist=speed *time
This is a pythag problem :a^2+b^=c^2, where
a=16t, bike dist
b=5(t+1), walking dist
c=52, dist apart in t hrs from 8 AM
(16t)^2+(5(t+1) ^2 =52^2
256t^2+(5t+5) ^2 =2704
256t^2+25t^2+50t+25=2704
281t^2+50t+25-2704=0
281t^2+50t-2679=0
solve this equation using the quadratic formula
positive solution t=3hrs
At 11Am they will be 52km apart
sartgakkamble:
very very thank you
ur welcome
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