Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10kmh-l. The second car goes at a speed of 8kmh-in the first hour and thereafter increasing the speed by 0.5kmh-1 each succeeding hour. After how many hours will the two cars meet?
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Two cars start together in the same direction from the same place. The first goes with uniform speed of 10km/h. The second goes at a speed of 8km/h in the first hour and increases the speed by
2
1
km in each succeeding hour. After how many hours will the second car overtake the first car if both cars go non-stop?
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Solution
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Suppose the second car overtakes the first car after t hours.
Then, the two cars travel the same distance in t hours.
Distance travelled by the first car in t hours =10tkm.
Distance travelled by the second car in t hours.
= Sum of t terms of an A.P. with first term 8 and common difference 1/2.
=
2
t
{2×8+(t−1)×
2
1
}=
4
t(t+31)
When the second car overtakes the first car, we have
10t=
4
t(t+31)
⟹40t=t
2
+31t⟹t
2
−9t=0
⟹t(t−9)=0⟹t=9 [∵t
=0]
Thus, the second car will overtake the first car in 9 hours.