An airplane is flying straight and level at a speed vo = 150 mph and with a constant time rate of increase of speed v' = 20 ft/s^2, when it starts to climb along a circular path with a radius of curvature p = 2000 ft. The airplane maintains v' constant for about 30s. Determine the acceleration of the airplane 25s after the start of the climb and express the result in the Cartersian component system shown.
To determine the acceleration of the airplane 25s after the start of the climb, we need to break it down into two components: tangential and radial.
1. Tangential acceleration (at):
The tangential acceleration can be found using the formula:
at = v' * t,
where v' is the rate of increase of speed and t is the time.
Given:
v' = 20 ft/s^2,
t = 25s.
Substituting the values into the formula, we get:
at = 20 ft/s^2 * 25s = 500 ft/s.
2. Radial acceleration (ar):
The radial acceleration can be found using the formula:
ar = v^2 / p,
where v is the instantaneous velocity and p is the radius of curvature.
Given:
v = vo + at * t,
vo = 150 mph (convert to ft/s),
at = 500 ft/s,
t = 25s,
p = 2000 ft.
Converting the initial velocity to ft/s:
vo = 150 mph * 5280 ft/mile * (1/3600) hour/s = 220 ft/s.
Substituting the values into the formula, we get:
v = 220 ft/s + 500 ft/s * 25s = 12720 ft/s.
Substituting the values into the formula for radial acceleration, we get:
ar = (12720 ft/s)^2 / 2000 ft = 81264 ft/s^2.
Therefore, the acceleration of the airplane 25s after the start of the climb is:
ar = 81264 ft/s^2 (radial component).
at = 500 ft/s (tangential component).
To determine the acceleration of the airplane 25s after the start of the climb, we can follow these steps:
Step 1: Convert the given speed vo from mph to ft/s:
vo = 150 mph = 150 * 5280 ft/3600 s = 220 ft/s
Step 2: Using the given time rate of increase of speed v' (velocity derivative), calculate the change in velocity in 30 seconds:
Δv = v' * t = 20 ft/s^2 * 30 s = 600 ft/s
Step 3: Calculate the final velocity vf after 30 seconds:
vf = vo + Δv = 220 ft/s + 600 ft/s = 820 ft/s
Step 4: Determine the centripetal acceleration ac using the formula:
ac = vf^2 / p
where p is the radius of curvature (2000 ft).
Substitute the values:
ac = (820 ft/s)^2 / 2000 ft = 672400 / 2000 ft/s^2 = 336.2 ft/s^2
Step 5: The total acceleration of the airplane can be broken down into two components: centripetal acceleration ac towards the center of the circular path and tangential acceleration at along the path. At this point, we need to find the tangential acceleration.
To find the tangential acceleration, we use the formula:
at = v' * t
where t is the time period for which the velocity derivative is constant (30 seconds).
Substitute the values:
at = 20 ft/s^2 * 30 s = 600 ft/s
Step 6: Now, we know the centripetal acceleration ac (336.2 ft/s^2) towards the center of the circular path and the tangential acceleration at (600 ft/s) along the path.
The acceleration of the airplane 25 seconds after the start of the climb can be represented in Cartesian component system as follows:
Horizontal component: ax = at = 600 ft/s
Vertical component: ay = ac = 336.2 ft/s^2
Hence, the acceleration of the airplane 25s after the start of the climb is approximately:
ax = 600 ft/s
ay = 336.2 ft/s^2