Question

Two cars leave a town at the same time and travel at constant speeds along straight roads that meet at an angle of 60° in the town. If one car travels twice as fast as the other and the distance between them increases at the rate of 81 mi/h, how fast is the slower car traveling? (Round your answer to the nearest integer.)

Answer 1

Make a sketch and use the cosine law:

let the rate of the slower car be x mph
then the rate of faster car is 2x mph

Let the distance covered by the slower car by x
then the distance covered by the faster car is 2x
let the distance between them be d
d^2 = x^2 + 4x^2 -2x(2x)cos60°
= 5x^2 - 2x^2
d^2 = 3x^2
d = √3 x
dd/dt = √3 dx/dt
81/√3 =dx/dt = appr 46.877 mph

slower car is appr 47 mph
faster car is appr 94 mph