A car traveling at 45 mph overtakes a cyclist who, riding at 18 mph, had a 2-hour head start. How far from the starting point does the car overtake the cyclist?

Answer 1

speed of car = 45 mph

speed of cyclist = 18 mph

Cyclist had a head start (early start) of 2 hours. In these 2 hours, cyclist already travelled 2x18 = 36miles

car cyclist meeting point

A------------36 miles--------B---------------------C

Now car is starting at point A and cyclist is already at point B which is 36 miles ahead. Since car is traveling faster than the cyclist, the car will catchup with the cyclist and overtake at point C.

We need to find AC.

AC = AB + BC = 36 + BC

Both car and cyclist start and travel for the same time until they meet. Let this time be 't' hours.

In 't' hours, car will travel 45 x t miles. Cyclist will travel 18 x t miles.

AC = 45t; BC = 18t.

We know AC = 36 + BC

so 45t = 36 + 18t. => 45t - 18t = 36. => 27t = 36 => t = 36/27 = 4/3 hours

So AC = 45t = 45 x 4/3 = 60 miles. So they both will meet at 60 miles from the car's starting point and at that point, the car will overtake the biker.

Answer 2

Cyclist going 18 mph went 36 miles in 120 minutes (2 hours).

Car traveling at 45 mph met cyclist's starting point at 36 miles after 48 minutes.

The car is traveling 2.5x faster than the cyclist, so the car would surpass the cyclist at mile 60 (36+24).

45 mph x h = 18 mph (2 + h)

h = 1.33

45 x 1.33 = 60 miles