6. What is the solution of the
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Let s = the speed of the slower train in km/hrs+10 = the speed of the faster train in km/hrLet t = time in hrs for the slower train to go 160 kmt-5 = time in hrs for faster train to go 160 km---------------Slower train:(1)
160=s.t
Faster train:(2) 160=(s+10) .(t-.5)From (1)(1) t=160/s
use the relations given in following screenshot and then use quadratic equation.
160=s.t
Faster train:(2) 160=(s+10) .(t-.5)From (1)(1) t=160/s
use the relations given in following screenshot and then use quadratic equation.