EXAM STYLE QUESTION
An aircraft is at rest at one end of a runway with its engine idling. The runway is 2.2km long. The aircraft then accelerates along the runway at an acceleration of 2.5ms-2 until it reaches its take off speed of 75ms-1
a)Calculate:
i)the time taken for the aircraft to reach its take-off speed (2 marks)
ii)the distance travelled in this time (2 marks)
b)As the aircraft reaches its takeoff speed, a warning light alerts the pilot to a possible fault in one of the engines.If the brakes are fully applied the maximum deceleration of the aircraft is 4.0ms-2. Before the pilot can take any action 2.5s have already elapsed, during which time the aircraft is continuing to fly at its take off speed. Determine whether the aircraft can be brought to rest within the remaining length of the runway? (5 marks)
Please help I am confused and need lots of clear detailled help with method all shown. Please dont abbreviate and use the actual words at all times
http://physics-ref.blogspot.com/2014/10/physics-9702-doubts-help-page-1.html
a)
i) To calculate the time taken for the aircraft to reach its take-off speed, we can use the formula:
time = (final velocity - initial velocity) / acceleration
Given that the initial velocity is 0 m/s, final velocity is 75 m/s, and acceleration is 2.5 m/s², we can substitute these values into the formula:
time = (75 - 0) / 2.5
time = 75 / 2.5
time = 30 seconds
Therefore, it takes 30 seconds for the aircraft to reach its take-off speed.
ii) To calculate the distance travelled by the aircraft in this time, we can use the formula:
distance = initial velocity × time + 0.5 × acceleration × time²
Since the initial velocity is 0 m/s and the acceleration is 2.5 m/s², we can substitute these values into the formula, along with the time of 30 seconds:
distance = 0 × 30 + 0.5 × 2.5 × (30)²
distance = 0 + 0.5 × 2.5 × 900
distance = 0 + 0.5 × 2.5 × 900
distance = 0 + 1125
distance = 1125 meters
Therefore, the distance travelled by the aircraft in this time is 1125 meters.
b) In order to determine if the aircraft can be brought to rest within the remaining length of the runway, we need to calculate the distance it will take for the aircraft to come to a stop when maximum deceleration is applied.
Using the formula for distance, we can set the final velocity to 0 m/s, initial velocity to 75 m/s, and acceleration to -4.0 m/s² (since deceleration is negative):
distance = (final velocity² - initial velocity²) / (2 × acceleration)
Plug in the values:
distance = (0² - 75²) / (2 × -4.0)
distance = (-5625) / (-8.0)
distance = 703.125 meters
So, when maximum deceleration is applied, it will take approximately 703.125 meters for the aircraft to come to a stop.
Since the remaining length of the runway is 2.2 km (or 2200 meters), we compare the calculated stopping distance of 703.125 meters with the remaining length:
703.125 < 2200
Since 703.125 is less than 2200, the aircraft can indeed be brought to rest within the remaining length of the runway.
Therefore, the aircraft can be brought to rest within the remaining length of the runway.
To solve this problem, we will use kinematic equations. These equations relate the position, velocity, time, and acceleration of an object.
a) i) The question asks for the time taken to reach the take-off speed.
We can use the equation of motion: v = u + at
Where:
v is the final velocity (take-off speed) = 75 m/s
u is the initial velocity (resting) = 0 m/s
a is the acceleration = 2.5 m/s^2
Rearranging the equation, we have:
t = (v - u) / a
Plugging in the values:
t = (75 - 0) / 2.5 = 30 seconds
Therefore, it takes 30 seconds for the aircraft to reach its take-off speed.
ii) The question asks for the distance traveled in this time.
We can use another equation of motion: s = ut + 1/2 at^2
Where:
s is the distance traveled
u is the initial velocity (resting) = 0 m/s
t is the time taken = 30 s
a is the acceleration = 2.5 m/s^2
Plugging in the values:
s = (0 × 30) + (1/2 × 2.5 × 30^2)
s = 0 + (1/2 × 2.5 × 900)
s = 1125 meters
Therefore, the distance traveled in this time is 1125 meters.
b) To determine if the aircraft can be brought to rest within the remaining length of the runway, we need to calculate the distance it will cover during the deceleration.
We can use the same equation of motion: s = ut + 1/2 at^2
Where:
s is the distance traveled
u is the initial velocity (take-off speed) = 75 m/s
t is the time taken for deceleration (unknown)
a is the deceleration = -4.0 m/s^2 (negative sign indicates deceleration)
We need to solve for t.
Rearranging the equation, we have:
t^2 + (2u/a)t - 2s/a = 0
Plugging in the values:
t^2 + (2 × 75 / -4)t - 2.2 km (-4.0) = 0
t^2 - 37.5t + 2.2 km = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values:
t = (-(-37.5) ± √((-37.5)^2 - 4 × 1 × 2.2 km)) / (2 × 1)
t = (37.5 ± √(1406.25 - 8.8 km)) / 2
Calculating the square root inside the square root:
√(1406.25 - 8.8 km)
= √(1406.25 - 8800)
= √(7406.25)
= 86.0026 m
Substituting this value back into the equation for t:
t = (37.5 ± 86.0026) / 2
There are two possible values for t:
t1 = (37.5 + 86.0026) / 2 = 123.5026 / 2 = 61.7513 s
t2 = (37.5 - 86.0026) / 2 = -48.5026 / 2 = -24.2513 s
Since time cannot be negative, we discard t2.
Therefore, the time taken for deceleration is approximately 61.7513 seconds.
Now, we need to calculate the distance covered during this deceleration. We can use the same equation of motion:
s = ut + 1/2 at^2
Where:
s is the distance traveled
u is the initial velocity (take-off speed) = 75 m/s
t is the time taken for deceleration = 61.7513 s
a is the deceleration = -4.0 m/s^2 (negative sign indicates deceleration)
Plugging in the values:
s = (75 × 61.7513) + (1/2 × -4.0 × 61.7513^2)
Calculating this value:
s = 4616.335 + (-7727.692)
s = -3111.357
The negative sign indicates that the object moved in the opposite direction (decelerated) from its initial motion.
Therefore, the distance covered during deceleration is approximately 3111.357 meters.
To determine if the aircraft can be brought to rest within the remaining length of the runway, we subtract the distance covered during the deceleration from the total length of the runway:
Remaining distance = 2.2 km - 3.111357 km (converted to meters)
Remaining distance = 2200 - 3111.357
Remaining distance = -911.357 meters
The negative value means that the aircraft does not have enough distance to come to a complete stop. It will overshoot the end of the runway.
Therefore, the aircraft cannot be brought to rest within the remaining length of the runway.