A tourist leaves point $A$ on a bicycle. Having travelled for $1.5$ h at $16$ km/h, he makes a stop for $1.5$ h and then pedals on with the same speed. Four hours after the start of the first tourist, a second tourist leaves point $A$ on a motor-cycle and rides with the speed of $56$ km/h in the same direction. What distance will they cover before the second tourist overtakes the first?

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Solution

Let time taken by second one to overtake First one be $x h r s$
Total distance travelled by second one is $56 x$
Meanwhile distance covered by First one is $1.5 (16) + 1.5 (0) + 16 (x + 1)$
$⟹ 16 x + 40$
at point of overtaking
$56 x = 16 x + 40$
$⟹ x = 1$
so they travel
$56 k m$ before second overtakes the first one..

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Let time taken by second one to overtake First one be xhrsTotal distance travelled by second one is 56xMeanwhile distance covered by First one is 1.5(16)+1.5(0)+16(x+1)⟹16x+40at point of overtaking56x=16x+40⟹x=1so they travel 56kmbefore second overtakes the first one..