At an air show, a jet plane has velocity components vx=695km/h and vy=435km/h at time 4.65 s and vx=868km/h and vy=365km/h at time 7.52 s.
a)For this time interval, find the x component of the plane's average acceleration.
Express your answer in kilometers per hour per second.
b)For this time interval, find the y component of the plane's average acceleration.
Express your answer in kilometers per hour per second.
c)For this time interval, find the magnitude of its average acceleration.
Express your answer in kilometers per hour per second to two significant figures.
d)For this time interval, find the direction of its average acceleration.
Express your answer in degrees.
a. a1 = (Vx2-Vx1)/(T2-T1) = (868-695)/(7.52-4.65) = 60.28 km/h/s.
b. a2 = (Vy2-Vy1)/(T2-T1) =
c. a = sqrt((a1)^2 + (a2)^2) =
d. TanA = a1/a2. The angle is measured CW from =+y-axis.
To find the average acceleration of the plane, we need to use the following formulas:
a) Average acceleration (ax) = (final velocity - initial velocity) / time interval
b) Average acceleration (ay) = (final velocity - initial velocity) / time interval
c) Magnitude of average acceleration = √(ax^2 + ay^2)
d) Direction of average acceleration = arctan(ay / ax)
Let's apply these formulas to solve the problem.
a) For the x-component average acceleration:
- Initial velocity (v0x) = 695 km/h
- Final velocity (vf_x) = 868 km/h
- Time interval (Δt) = 7.52 s - 4.65 s = 2.87 s
Using the formula, ax = (vf_x - v0x) / Δt, we get:
ax = (868 km/h - 695 km/h) / 2.87 s = 60.2 km/(h·s)
Therefore, the x-component of the plane's average acceleration is 60.2 km/(h·s).
b) For the y-component average acceleration:
- Initial velocity (v0y) = 435 km/h
- Final velocity (vf_y) = 365 km/h
- Time interval (Δt) = 7.52 s - 4.65 s = 2.87 s
Using the formula, ay = (vf_y - v0y) / Δt, we get:
ay = (365 km/h - 435 km/h) / 2.87 s = -103.8 km/(h·s)
Note: The negative sign indicates a downward direction.
Therefore, the y-component of the plane's average acceleration is -103.8 km/(h·s).
c) To find the magnitude of the average acceleration, we use the formula mentioned above:
Magnitude of average acceleration = √(ax^2 + ay^2) = √((60.2 km/(h·s))^2 + (-103.8 km/(h·s))^2)
Calculating the magnitude, we get:
Magnitude of average acceleration ≈ 120.3 km/(h·s)
Therefore, the magnitude of the plane's average acceleration is approximately 120.3 km/(h·s).
d) To find the direction of the average acceleration, we use the formula mentioned above:
Direction of average acceleration = arctan(ay / ax) = arctan((-103.8 km/(h·s)) / (60.2 km/(h·s)))
Calculating the direction, we get:
Direction of average acceleration ≈ -59.6 degrees
Therefore, the direction of the plane's average acceleration is approximately -59.6 degrees.