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}](https://tex.z-dn.net/?f=d%5E%7B2%7D%3Dh%5E%7B2%7D%2Bx%5E%7B2%7D)
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}}](https://tex.z-dn.net/?f=x%3D%5Csqrt%7Bd%5E%7B2%7D-h%5E%7B2%7D%7D)
and substitute the given values:
![x=\sqrt{(1)^{2}-(0.5)^{2}}](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B%281%29%5E%7B2%7D-%280.5%29%5E%7B2%7D%7D)
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}](https://tex.z-dn.net/?f=d%5E%7B2%7D%3D%280.5%29%5E%7B2%7D%2Bx%5E%7B2%7D)
![d^{2}=0.25+x^{2}](https://tex.z-dn.net/?f=d%5E%7B2%7D%3D0.25%2Bx%5E%7B2%7D)
Next, we can go ahead and differentiate the equation so we get:
![2d\frac{dd}{dt}=2x\frac{dx}{dt}](https://tex.z-dn.net/?f=2d%5Cfrac%7Bdd%7D%7Bdt%7D%3D2x%5Cfrac%7Bdx%7D%7Bdt%7D)
which can be simplified to:
![d\frac{dd}{dt}=x\frac{dx}{dt}](https://tex.z-dn.net/?f=d%5Cfrac%7Bdd%7D%7Bdt%7D%3Dx%5Cfrac%7Bdx%7D%7Bdt%7D)
we can next solve the equation for dx/dt so we get:
![\frac{dx}{dt}=\frac{d}{x}*\frac{dd}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bx%7D%2A%5Cfrac%7Bdd%7D%7Bdt%7D)
so we can now substitue te provided values so we get:
![\frac{dx}{dt}=\frac{1}{0.866}*-190](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B0.866%7D%2A-190)
so we get:
![\frac{dx}{dt}=-219.40mph](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%3D-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}](https://tex.z-dn.net/?f=V_%7Bch%7D%3DV_%7Bc%7D-V_%7Bh%7D)
so we can solve this for the actual velocity of the car, so we get:
![V_{c}=V_{ch}+V_{h}](https://tex.z-dn.net/?f=V_%7Bc%7D%3DV_%7Bch%7D%2BV_%7Bh%7D)
so we get:
![V_{c}=-219.40mph+150mph](https://tex.z-dn.net/?f=V_%7Bc%7D%3D-219.40mph%2B150mph)
which yields:
![V_{c}=-69.4mph](https://tex.z-dn.net/?f=V_%7Bc%7D%3D-69.4mph)
So it has a velocity of 69.4mph in a direction oposite to the helicopter's movement.