Two are craft P and Q are flying at the same speed, 300m/sec. The direction along which Pis flying 15 at right angle 70 the direction along which Q is flying find the magnitude of the velocity of the aircraft P relative to aircraft Q 240 cm/hr
what does
"15 at right angle 70"
even mean?
and a relative speed in cm/hr? Really? I think you should repost this using the actual language. It makes no sense as given here.
To find the magnitude of the velocity of aircraft P relative to aircraft Q, we need to use vector addition.
Let's break down the velocity vectors of P and Q into their respective components.
The magnitude of the velocity of aircraft P is given as 300 m/s.
The direction along which P is flying is at a right angle of 70 degrees from the reference direction (let's assume it's the positive x-axis).
The component of the velocity of aircraft P in the x-direction would be given by:
P_x = P * cos(angle)
= 300 * cos(70 degrees)
≈ 96.03 m/s
The component of the velocity of aircraft P in the y-direction would be given by:
P_y = P * sin(angle)
= 300 * sin(70 degrees)
≈ 289.07 m/s
Similarly, we can find the components of the velocity of aircraft Q.
Since the magnitude of the velocity of aircraft Q is not given, we'll denote it as Q.
The direction along which Q is flying is at a right angle of 15 degrees from the reference direction.
The component of the velocity of aircraft Q in the x-direction would be given by:
Q_x = Q * cos(angle)
= Q * cos(15 degrees)
The component of the velocity of aircraft Q in the y-direction would be given by:
Q_y = Q * sin(angle)
= Q * sin(15 degrees)
Now, we need to find the difference in the velocity vectors of P and Q (P - Q) to determine the relative velocity vector.
The component of the relative velocity vector in the x-direction would be:
(P - Q)_x = P_x - Q_x
= 96.03 m/s - Q * cos(15 degrees)
The component of the relative velocity vector in the y-direction would be:
(P - Q)_y = P_y - Q_y
= 289.07 m/s - Q * sin(15 degrees)
The magnitude of the velocity of aircraft P relative to aircraft Q can be found using the Pythagorean theorem:
Magnitude = sqrt((P - Q)_x^2 + (P - Q)_y^2)
Given that the magnitude is 240 cm/hr, we need to convert it to a consistent unit (m/s).
240 cm/hr = (240/100) m/hr = (240/100) * (1/3600) m/s ≈ 0.0067 m/s
Now we can set up the equation:
0.0067 = sqrt((96.03 - Q * cos(15 degrees))^2 + (289.07 - Q * sin(15 degrees))^2)
To solve this equation for Q, we need to use numerical methods or an equation solver as it cannot be solved algebraically.
The value of Q can be found by numerical methods or by using an equation solver tool like MATLAB or Wolfram Alpha.
To find the magnitude of the velocity of aircraft P relative to aircraft Q, we need to subtract the velocity vector of Q from the velocity vector of P.
Given:
- The speed of both aircraft P and Q is 300 m/sec.
- The direction along which P is flying makes a 15-degree angle with the right angle (90 degrees).
- The direction along which Q is flying makes a 70-degree angle with the right angle (90 degrees).
- The given magnitude of the relative velocity is 240 cm/hr.
To solve this problem, we can break down the velocity vectors of P and Q into their horizontal and vertical components. The horizontal component represents the velocity in the x-axis direction, while the vertical component represents the velocity in the y-axis direction.
First, we need to convert the magnitude of the relative velocity from centimeters per hour (cm/hr) to meters per second (m/sec). Since 1 meter = 100 centimeters and 1 hour = 3600 seconds, we can use the following conversion:
240 cm/hr = (240/100) m/sec = 2.4 m/sec
Now, let's break down the velocity vector of P into its horizontal and vertical components. Since P is flying at a right angle of 15 degrees, we can use trigonometry to find these components:
Horizontal component of P's velocity:
Vp_horizontal = Vp * cos(angle)
Vp_horizontal = 300 m/sec * cos(15 degrees)
Vertical component of P's velocity:
Vp_vertical = Vp * sin(angle)
Vp_vertical = 300 m/sec * sin(15 degrees)
Similarly, we break down the velocity vector of Q into its horizontal and vertical components:
Horizontal component of Q's velocity:
Vq_horizontal = Vq * cos(angle)
Vq_horizontal = 300 m/sec * cos(70 degrees)
Vertical component of Q's velocity:
Vq_vertical = Vq * sin(angle)
Vq_vertical = 300 m/sec * sin(70 degrees)
Now, we have the horizontal and vertical components for both P and Q. To find the magnitude of the velocity of P relative to Q, we subtract the corresponding components:
Velocity relative (Vrelative) = [(Vp_horizontal - Vq_horizontal)^2 + (Vp_vertical - Vq_vertical)^2]^0.5
Vrelative = [(Vp_horizontal - Vq_horizontal)^2 + (Vp_vertical - Vq_vertical)^2]^0.5
Plug in the previously calculated values for Vp_horizontal, Vp_vertical, Vq_horizontal, and Vq_vertical into this equation to find the magnitude of the relative velocity.