A small, single engine airplane is about to take off. The airplane becomes airborne, when its speed reaches 117.0 km/h. The conditions at the airport are ideal, there is no wind. When the engine is running at its full power, the acceleration of the airplane is 2.40 m/s2. What is the minimum required length of the runway?
I thought you used the equation: Vf^2=Vi^2+2ad where Vi is 0 but I can't seem to get the right answer. Can someone help please?!?
V = 117km/h = 117000m/3600s. = 32.5 m/s.
V^2 = Vo^2 + 2a*L = 32.5^2
0 + 4.8*L = 1056.25
L = 1056.25/4.8 = 220 m.
V = Vo + a*t = 32.5 m/s.
0 + 2.4*
V = 117km/h = 117000m/3600s. = 32.5 m/s.
V^2 = Vo^2 + 2a*L = 32.5^2
0 + 4.8*L = 1056.25
L = 1056.25/4.8 = 220 m.
Yes, you are correct that the equation to use is the equation of motion, which is:
Vf^2 = Vi^2 + 2ad
Let's break down the problem and calculate the minimum required length of the runway step-by-step.
Given:
Vi (initial velocity) = 0 km/h (since the airplane is at rest initially)
Vf (final velocity) = 117.0 km/h = 117.0 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 32.5 m/s
a (acceleration) = 2.40 m/s^2
We want to find d (distance).
Now we can substitute the known values into the equation:
32.5^2 = 0 + 2 * 2.40 * d
1056.25 = 4.8d
Divide both sides of the equation by 4.8:
d = 1056.25 / 4.8 = 220.14 meters
Therefore, the minimum required length of the runway is approximately 220.14 meters.
To find the minimum required length of the runway, we can use the equation you mentioned: Vf^2 = Vi^2 + 2ad. Let's break down the problem step by step.
First, we need to determine the initial velocity (Vi) of the airplane. In this case, the airplane is just starting to take off and becomes airborne when its speed reaches 117.0 km/h. To convert this speed into m/s, we divide it by 3.6 (since 1 km/h = 1/3.6 m/s).
So, Vi = 117.0 km/h / 3.6 = 32.5 m/s (approximately, rounding to one decimal place).
Next, we need to find the acceleration (a) of the airplane, which is given as 2.40 m/s^2.
Now we can substitute these values into the equation: Vf^2 = Vi^2 + 2ad.
Since the airplane becomes airborne, the final velocity (Vf) is also 0 m/s.
0^2 = (32.5)^2 + 2 * 2.40 * d (where d represents the minimum required length of the runway)
Simplifying the equation:
0 = 1056.25 + 4.80d
Rearranging the equation:
4.80d = -1056.25
Finally, we can solve for d by dividing both sides of the equation by 4.80:
d = -1056.25 / 4.80
d ≈ -220.05
Since distance cannot be negative, we can disregard the negative sign.
Therefore, the minimum required length of the runway is approximately 220.05 meters.