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.)
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
To find the speed of the slower car, we can set up a system of equations using the given information. Let's call the speed of the slower car "x" miles per hour.
Since the other car travels twice as fast as the slower car, its speed would be 2x miles per hour.
We are given that the distance between the two cars increases at a rate of 81 mi/h. This means that the rate of change of the distance with respect to time is 81 mi/h.
Using trigonometry, we can relate the rate of change of the distance to the speeds of the cars. The rate of change of the distance can be expressed as the sum of the rates at which the two cars move along their respective roads.
The rate at which the slower car moves is x mi/h, and the rate at which the faster car moves is 2x mi/h. Since the roads meet at an angle of 60°, we can use the equation:
x + 2x*cos(60°) = 81
Simplifying this equation:
x + 2x*0.5 = 81
3x = 81
Now, we can solve for x by dividing both sides of the equation by 3:
x = 81 / 3
x = 27
Therefore, the speed of the slower car is 27 miles per hour.