Answer :
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.
[tex]d^{2}=h^{2}+x^{2}[/tex]
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:
[tex]x=\sqrt{d^{2}-h^{2}}[/tex]
and substitute the given values:
[tex]x=\sqrt{(1)^{2}-(0.5)^{2}}[/tex]
which yields:
x=0.866mi
we know that the height of the helicopter is going to be constant, so we rewrite the equation as:
[tex]d^{2}=(0.5)^{2}+x^{2}[/tex]
[tex]d^{2}=0.25+x^{2}[/tex]
Next, we can go ahead and differentiate the equation so we get:
[tex]2d\frac{dd}{dt}=2x\frac{dx}{dt}[/tex]
which can be simplified to:
[tex]d\frac{dd}{dt}=x\frac{dx}{dt}[/tex]
we can next solve the equation for dx/dt so we get:
[tex]\frac{dx}{dt}=\frac{d}{x}*\frac{dd}{dt}[/tex]
so we can now substitue te provided values so we get:
[tex]\frac{dx}{dt}=\frac{1}{0.866}*-190[/tex]
so we get:
[tex]\frac{dx}{dt}=-219.40mph[/tex]
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:
[tex]V_{ch}=V_{c}-V_{h}[/tex]
so we can solve this for the actual velocity of the car, so we get:
[tex]V_{c}=V_{ch}+V_{h}[/tex]
so we get:
[tex]V_{c}=-219.40mph+150mph[/tex]
which yields:
[tex]V_{c}=-69.4mph[/tex]
So it has a velocity of 69.4mph in a direction oposite to the helicopter's movement.
The rate of change of the distance between the car and the helicopter is
a function of the speed of the car.
Response:
- The speed of the car is approximately 69.4 mph
(a) The drawing of the situation created with MS Visio is attached
(b) The given quantities are;
- Altitude; 0.5 m
- Speed of the helicopter; 150 mph
- Distance between the car and the helicopter; 1 mile
- Rate of change of the distance between the helicopter and the car, [tex]\dfrac{dr}{dt}[/tex] = -190 mph
Which method is used to find the speed of the car?
Given:
Speed of the helicopter = 150 mph
Altitude of the helicopter = 0.5 mile
Distance of the oncoming car, r = 1 mile
Rate at which the distance between the helicopter and the car is decreasing = 190 mph
Where;
Horizontal distance of the car, x = √((1 mile)² - (0.5 mile)²) = √(0.75) mile
According to Pythagorean theorem, we have; r² = 0.5² + x²
The instantaneous rate of change of the distance between the car and the helicopter is changing is; [tex]\mathbf{\dfrac{dr}{dt} } = -190[/tex]
Differentiating r² = 0.5² + x² gives;
[tex]\dfrac{d}{dt} r^2 = \mathbf{\dfrac{d}{dt} (0.5^2 + x^2)}[/tex]
[tex]2 \cdot r \cdot \dfrac{dr}{dt} = \mathbf{ 2 \cdot x \cdot \dfrac{dx}{dt}}[/tex]
Plugging in the values, we get;
[tex]2 \times 1 \times (-190)= 2 \times \sqrt{0.75} \cdot \dfrac{dx}{dt}[/tex]
[tex]\dfrac{dx}{dt} = \dfrac{2 \times 1 \times (-190) }{2 \times \sqrt{0.75}} \approx \mathbf{-219.4}[/tex]
[tex]\left| \dfrac{dx}{dt} \right| = \mathbf{\left|v_1 \right| + \left|v_2 \right|}[/tex]
Where;
v₂ = The velocity of the car
Which gives;
[tex]\left|v_2 \right| \approx 219.4 -150 = \mathbf{69.4}[/tex]
- The speed of the car, v₂ ≈ 69.4 mph
(a) Please find attached the drawing of the situation created with MS Visio
(b) The given quantities are;
- Altitude of the helicopter = 0.5 m
- Speed of the helicopter, v₁ = 150 mph
- Distance between the car and the helicopter, r = 1 mile
- Rate of change of the distance between the helicopter and the car, [tex]\dfrac{dr}{dt}[/tex] = -190 mph
Learn more about instantaneous rate of change here:
https://brainly.com/question/19741268
https://brainly.com/question/1407601
