Two cars are at a red light, just at the beginning of a 90 left turn on the road. The cars are next to each other apart, and they start exactly at the same time when the light turns green. The car on the inside of the turn makes a quarter circle of radius while the car on the outside makes the turn on a circle of radius The centers of the turning circles coincide.
As they accelerate in their turns, they keep going head-to-head, i.e. the line connecting the drivers' heads keeps pointing to the common center of the turning circles. Then they reach the straight section of the road at the same time, with each car stopping accelerating and moving with constant velocity afterwards.
Where is the outside car at the moment when the inside car is out of the turn?
Note: The illustration is not drawn to scale.
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In a time interval of Δ t the cars move a distance of Δ α R 1 and Δ α R 2 , where R is the radius of the turning circle and Δ α is the change in the angle of the line connecting the driver's head, measured at the center of the turning circle. Using v 1 = R 1 Δ α / Δ t and v 2 = R 2 Δ α / Δ t and dividing the two equations we get v 2 / v 1 = R 2 / R 1 This relationship holds at any moment while they are in the turn. In particular, it holds at the moment when they exit the turn, and determines the ratio of the velocities after the turn. After exiting the turn the distance traveled by each car is s 1 = v 1 t and s 2 = v 2 t , therefore s 2 / s 1 = v 2 / v 1 . The distance between the cars is s 2 − s 1 = ( v 2 / v 1 − 1 ) s 1 = ( R 2 / R 1 − 1 ) s 1 = 1 2 . 5 m
It is interesting that the problem can be solved without knowing the actual velocities or accelerations. In fact, the cars do not have to accelerate uniformly; the only requirement is that they travel head to head during the turn.