Two cars started from same point at the same time, moving at constant speeds. If one car passed the other 1 km from start, where will be the second passing?
The two cars on a circuit puzzle
Two friends took their cars to a nearby circuitous route and started driving at the same time from same point S. The circuit was longer than 11 km, both cars moved at constant speeds, but one moved faster than the other.
If the faster car passed the second car first time at a point 1 km from the starting point in the direction of their movement, at what point will be the second passing?
Time to solve 5 minutes.
In essence, it is a race between two objects moving at different constant speeds in same direction on a circuit.
Solution to the two cars on a circuitous route puzzle
First urge is to do the race math as you may be experienced in race math. But as you analyze the puzzle more, realization dawns that variables are too many. It is not a straightforward case of mathematical deductions.
Thinking of alternate course of action, it doesn't take long to discover the simple logic:
After all, what is the difference of status between the first starting point and the first meeting point?
Answer is immediate:
The car positions and speeds are exactly same at both points, as if the cars are starting a second time 1 km further down the route.
Solution:
The second passing will be exactly 1 km further down the route from the first passing point. This will be 2 kms from the starting point S.
Advantage and power of math reasoning: bypassing troublesome math.
But then, are you interested more in math rather than math reasoning? In that case, here is some math.
Mathematical solution of the two cars running at constant speeds puzzle
Assume speeds are S1 and S2 (both in km/hr) with S1 larger than S2 and length of the circuit D km.
Relative speed of the two cars is (S1 - S2) km/hr. At these speeds, the first car will move ahead by an increasing distance while both cars go on completing their laps of the circuit.
When the first car passes the second car for the first time 1 km further down the route from the starting point, virtually the second car stood still at the starting point and the first car moved ahead from the first by 1 km at a speed of (S1 - S2) km/hr.
If the time taken from the start to the first crossing is T hrs,
(S1 - S2) x T = 1, all speeds in km/hr, time in hr and distance in km.
Both cars moving at constant speeds, the second passing will also take exactly the same time T hr from the time of the first passing.
Relative speeds also being same, the relative distance by which the first car will be further ahead than the second car will be,
(S1 - S2) x T = 1 km again, but this time 1 km from the first passing point in the direction of movement.
Result is same:
The second passing will be exactly 2 kms from the starting point.
If you are still not satisfied and want some concrete figure for the variables, here it goes.
A mathematically correct example of the two cars running at constant speeds
In this examples, all the variables are in the form of numeric values that satisfy the puzzle conditions.
First car speed 61 km/hr, second car speed 49 km/hr and circuit length 12 km.
In 1 hour, the first car travels:
61 = 5 x 12 + 1, 5 circuit laps plus 1 km more.
In this 1 hour, the second car travels:
49 = 4 x 12 + 1, 4 circuit laps plus 1 km more.
What really happens:
With higher speed, in 1 hour, the first car moves ahead continuously and completes the five laps plus 1 km further down. In the same time the second car finally falls behind by exactly 1 lap but covers another 1 km down the route for the first passing.
In the second passing, the same events will be repeated, but with the starting point as the point of first passing.
This is not a mathematical proof.
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