The radar stations A and B, separated by the distance a = 500 m, track the plane C by recording the angles α and β at 1-second intervals,
If three successive readings are shown in the table,calculate the speed v of the plane and the climb angle y at t = 10 seconds as accurate as you can.
The coordinates of the plane can be shown to be:
x = a ((tan β)/(tan β – tan α)
y = a ((tan α tan β)/(tan β – tan α) (height to plane)
t/sec 9 10 11
α (degrees) 54.80 54.06 53.34
β (degrees) 65.59 64.59 63.62
To calculate the speed v of the plane and the climb angle y at t=10 seconds, we will use the given coordinates of the plane (α and β) and the formula:
x = a(tan β) / (tan β – tan α)
y = a(tan α tan β) / (tan β – tan α)
where a is the distance between radar stations A and B (500 m).
First, let's calculate the values of x and y at t=10 seconds:
For x:
- α at t=10 seconds: 54.06 degrees
- β at t=10 seconds: 64.59 degrees
x = 500 * (tan 64.59) / (tan 64.59 - tan 54.06)
x ≈ 850.67 meters
For y:
- α at t=10 seconds: 54.06 degrees
- β at t=10 seconds: 64.59 degrees
y = 500 * (tan 54.06 * tan 64.59) / (tan 64.59 - tan 54.06)
y ≈ 367.41 meters
Now, let's calculate the speed v at t=10 seconds:
To calculate the speed, we need to use the change in position (x) between t=9 seconds and t=10 seconds. So, we'll use the coordinates at t=9 seconds and t=10 seconds:
For x:
- x at t=9 seconds: 850.67 meters
- x at t=10 seconds: 500 * (tan 65.59) / (tan 65.59 - tan 54.06)
≈ 856.04 meters
Δx = x(t=10) - x(t=9)
= 856.04 - 850.67
≈ 5.37 meters
To calculate the time difference Δt:
Δt = 10 - 9
= 1 second
Now we can calculate the speed:
v = Δx / Δt
= 5.37 / 1
≈ 5.37 m/s
Finally, to calculate the climb angle y at t=10 seconds:
y = a(tan α tan β) / (tan β - tan α)
= 500 * (tan 54.06 * tan 64.59) / (tan 64.59 - tan 54.06)
≈ 367.41 meters
Therefore, the speed v of the plane at t=10 seconds is approximately 5.37 m/s, and the climb angle y is approximately 367.41 meters.