Question

A police helicopter is flying at 150 mph at a constant altitude of 0.5 mile above a straight road. The pilot uses radar to determine that an oncoming car is at a distance of exactly 1 mile from the helicopter, and that this distance is decreasing at 190 mph. Find the speed of the car. (a) Draw a picture of the situation for any time t. (b) What quantities are given in the problem

Answer 1

Answer:

a) (see attached picture)

b) d, h, velocity of helicopter, dd/dt

in the direction oposite to the movement of the helicopter.

Step-by-step explanation:

In order to solve this problem we must start by drawing a diagram that will represent the situation (see attached picture). As you may see, there is a right triangle being formed by the helicopter, the car and the graound. We can use this to build the equation we are going to use to model the problem.

$d^{2}=h^{2}+x^{2}$

we can use this equation to find the value of x at that very time, so we can do that by solving the equation for x, so we get:

$x=\sqrt{d^{2}-h^{2}}$

and substitute the given values:

$x=\sqrt{(1)^{2}-(0.5)^{2}}$

which yields:

x=0.866mi

we know that the height of the helicopter is going to be constant, so we rewrite the equation as:

$d^{2}=(0.5)^{2}+x^{2}$

$d^{2}=0.25+x^{2}$

Next, we can go ahead and differentiate the equation so we get:

$2d\frac{dd}{dt}=2x\frac{dx}{dt}$

which can be simplified to:

$d\frac{dd}{dt}=x\frac{dx}{dt}$

we can next solve the equation for dx/dt so we get:

$\frac{dx}{dt}=\frac{d}{x}*\frac{dd}{dt}$

so we can now substitue te provided values so we get:

$\frac{dx}{dt}=\frac{1}{0.866}*-190$

so we get:

$\frac{dx}{dt}=-219.40mph$

the negative sign means that the x-value is decreasing.

now, this problem deals with relative velocities, so we get that:

Velocity of car about the helicopter = Velocity of the car - Velocity of the helicopter. Or:

$V_{ch}=V_{c}-V_{h}$

so we can solve this for the actual velocity of the car, so we get:

$V_{c}=V_{ch}+V_{h}$

so we get:

$V_{c}=-219.40mph+150mph$

which yields:

$V_{c}=-69.4mph$

So it has a velocity of 69.4mph in a direction oposite to the helicopter's movement.