A jet plane has a takeoff speed of Vto=125km/h (34.72 m/s) and it's engines can power it to accelerate with an average acceleration of 5.6 m/s^2.
A.) what length of runway will it need to take off safely?
B.) suppose it had an available runway length of only 75m. What minimum constant acceleration would it need to take off safely?
C.) supposing the runway length and acceleration in question b, how long will the plane take to take off?
D.) upon arriving at its destination the plane lands with a speed of 500 km/h and it's brakes can cause it to accelerate at -4.5 m/s^2 without severely compromising the comfort of the passengers. How long will the airplane take to come to rest?
E.) suppose the runway is 1000 yards long. Will the plane be able to land safely?
Vo = 0
V = 34.72 m/s.
a = 5.6 m/s^2
A. V = Vo + a*t = 34.72
0 + 5.6t = 34.72
t = 6.2 s.
Length = Vo*t + 0.5a*t^2
Length = 0 + 2.8*6.2^2 = 107.6 m.
B. V^2 = Vo^2 + 2a*L = 34.72^2 = 1205.5
0 + 2a*75 = 1205.5
150a = 1205.5
a = 8.04 m/s^2.
C. V = Vo + a*t = 34.72 m/s.
0 + 8.04t = 34.72
t = 4.32 s.
D. Vo=500km/h = 500,000m/3600s=138.9 m/s
= Initial velocity.
V = Vo + a*t = 0
138.9 - 4.5t = 0
4.5t = 138.9
t = 30.86 s.
E. L=1000yds * 3ft/yd * 1m/3.3Ft=909.1 m
d=(V^2-Vo^2)/2a = (0-138.9^2)/-9=2144 m.
= Stopping distance.
No, the plane will not be able to land safely.
A.) To calculate the length of the runway the jet plane will need to take off safely, we can use the following kinematic equation:
v^2 = v0^2 + 2ad
Where:
v = final velocity = takeoff speed = 34.72 m/s
v0 = initial velocity = 0 m/s
a = acceleration = 5.6 m/s^2
d = distance (runway length)
Rearranging the equation to solve for d, we have:
d = (v^2 - v0^2) / (2a)
Plugging in the values, we can calculate the length of the runway:
d = (34.72^2 - 0^2) / (2 x 5.6) = 216.52 meters
Therefore, the jet plane will need a runway length of approximately 216.52 meters to take off safely.
B.) If the available runway length is only 75 meters, we can rearrange the same kinematic equation to solve for the minimum constant acceleration required:
a = (v^2 - v0^2) / (2d)
Plugging in the values, we have:
a = (34.72^2 - 0^2) / (2 x 75) = 12.98 m/s^2
Therefore, the jet plane would need a minimum constant acceleration of approximately 12.98 m/s^2 to take off safely with a runway length of 75 meters.
C.) To calculate the time it will take for the plane to take off, we can use the following kinematic equation:
v = v0 + at
Where:
v = final velocity = takeoff speed = 34.72 m/s
v0 = initial velocity = 0 m/s
a = acceleration = 12.98 m/s^2 (from part B)
t = time
Rearranging the equation to solve for t, we have:
t = (v - v0) / a
Plugging in the values, we can calculate the time:
t = (34.72 - 0) / 12.98 = 2.68 seconds
Therefore, the jet plane will take approximately 2.68 seconds to take off.
D.) To calculate the time it will take for the airplane to come to rest upon landing, we can use the following kinematic equation:
v = v0 + at
Where:
v = final velocity = 0 m/s (since it comes to rest)
v0 = initial velocity = 500 km/h = 138.89 m/s
a = deceleration = -4.5 m/s^2
t = time
Rearranging the equation to solve for t, we have:
t = (v - v0) / a
Plugging in the values, we can calculate the time:
t = (0 - 138.89) / -4.5 = 30.86 seconds
Therefore, the airplane will take approximately 30.86 seconds to come to rest.
E.) To determine if the plane will be able to land safely on a runway that is 1000 yards long, we first need to convert the runway length to meters:
1 yard = 0.9144 meters
Therefore, the runway length is:
1000 yards * 0.9144 meters/yard = 914.4 meters
Comparing this with the minimum runway length required for takeoff (from part A), which is approximately 216.52 meters, we can see that the runway length of 914.4 meters is longer than the minimum required length. Therefore, the plane will be able to land safely on the given runway.
A.) To find the length of runway the jet plane will need to take off safely, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (takeoff speed) = 34.72 m/s
u = initial velocity = 0 (since the plane starts from rest)
a = acceleration = 5.6 m/s^2
s = distance (length of runway)
Substituting the given values into the equation, we can solve for s:
(34.72)^2 = 0^2 + 2(5.6)s
1194.98 = 11.2s
s = 1194.98 / 11.2 ≈ 106.92 meters
Therefore, the length of the runway needed for the plane to take off safely is approximately 106.92 meters.
B.) If the available runway length is only 75 meters, we can use the same equation of motion to find the minimum constant acceleration the plane would need:
(34.72)^2 = 0^2 + 2(a)(75)
1203.78 = 150a
a ≈ 8.02 m/s^2
Therefore, the minimum constant acceleration the plane would need to take off safely with a runway length of 75 meters is approximately 8.02 m/s^2.
C.) To find how long the plane will take to take off with a runway length of 75 meters and the acceleration found in question B, we can use the equation of motion:
v = u + at
where:
v = final velocity (takeoff speed) = 34.72 m/s
u = initial velocity = 0 (since the plane starts from rest)
a = acceleration = 8.02 m/s^2
t = time
Substituting the given values into the equation, we can solve for t:
34.72 = 0 + (8.02)t
t ≈ 4.33 seconds
Therefore, the plane will take approximately 4.33 seconds to take off with a runway length of 75 meters and an acceleration of 8.02 m/s^2.
D.) To find how long the airplane will take to come to rest after landing, we can use the equation of motion:
v = u + at
where:
v = final velocity = 0 (since the airplane comes to rest)
u = initial velocity = 500 km/h = 500 * (1000/3600) m/s ≈ 138.89 m/s
a = acceleration = -4.5 m/s^2 (negative sign indicates deceleration)
t = time
Substituting the given values into the equation, we can solve for t:
0 = 138.89 - (4.5)t
t ≈ 30.86 seconds
Therefore, the airplane will take approximately 30.86 seconds to come to rest after landing.
E.) To determine if the plane will be able to land safely on a runway length of 1000 yards, we need to convert the runway length to meters:
1 yard = 0.9144 meters
1000 yards = 914.4 meters
Since the length of the runway is greater than the distance required for the plane to take off safely (106.92 meters), the plane will be able to land safely on a 1000 yard long runway.