Car Race Puzzle
Three cars are driving on a track that forms a perfect circle and is wide enough that multiple cars can pass anytime. The car that is leading in the race right now is driving at 55 MPH and the car that is trailing at the last is going at 45 MPH. The car that is in the middle is somewhere between these two speeds.
Right now, you can assume that there is a distance of x miles between the leading car and the middle car and x miles between the middle car and the last car and also, x is not equal to 0 or 1.
The cars maintain their speed till the leading car catches up with the last car and then every car stops. In this scenario, do you think of any point when the distance between any two pairs will again be x miles i.e. the pairs will be x distance apart at the same time ?
Yes, there will be a point in time when the distance between any two pairs of cars will again be x miles.
Let’s assume that the track has a circumference of C miles. When the leading car catches up with the last car, it will have completed one full lap of the track, and the other two cars will have completed some fraction of a lap. Let’s call this fraction f, where 0 < f < 1.
At this point, the leading car is at the starting point, and the other two cars are x miles apart on the track. The middle car is fC miles ahead of the trailing car, and (1-f)C miles behind the leading car. The trailing car is x + fC miles behind the leading car, and (1-f)C – x miles ahead of the middle car.
As the cars continue to move, the distance between the leading and middle car will decrease at a rate of 55 – v miles per hour, where v is the speed of the middle car. Similarly, the distance between the middle and trailing car will increase at a rate of v – 45 miles per hour.
There will be a point in time when the distance between the leading and middle car, and the distance between the middle and trailing car, are both equal to x. Let’s call this time t. At time t, the distance between the leading car and the middle car will be x, which means that the middle car will be (1-f)x miles behind the leading car. The distance between the middle car and the trailing car will also be x, which means that the trailing car will be x + fC – (1-f)x = fC + x miles behind the leading car.
We can solve for t by setting the distances equal to x and solving for v. We get:
55t – vx = 55x vx – 45t = 45x
Solving for v and equating the two expressions, we get:
v = (55x + 45t) / (t + x)
Substituting this expression for v into the equation for the distance between the middle and trailing car, we get:
fC + x = vt – 45t = (55x + 45t) / (t + x) * t – 45t
Solving for t, we get:
t = C / (10 + 2f)
This is the time at which the distance between any two pairs of cars will again be x miles. Note that this time will be different for different values of f, but there will always be a time when the distance between the cars is x.
Your Answer
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