1.A plane traveling at 300.0 km/h attempts to land on a 500 m runway. The plane’s engines and brakes accelerate uniformly at –5.0 m/s2. Will the plane be able to land safely? Show all your work. Include any equations used, the given and unknown. Show all calculations. Express numerical answers with the correct number of significant digits. Remember to carry one or two extra decimal places throughout calculation. Do not round until the end.
To determine whether the plane will land safely on the runway, we will need to calculate the time it takes for the plane to come to a stop.
Given:
Initial speed (vi) = 300.0 km/h
Acceleration (a) = -5.0 m/s^2
Distance to stop (d) = 500 m
First, let's convert the initial speed from km/h to m/s:
vi = 300.0 km/h * (1000 m / 1 km) * (1 h / 3600 s)
vi = 83.33 m/s (rounded to two decimal places)
Now, we can use the kinematic equation to find the time it takes for the plane to stop:
vf^2 = vi^2 + 2ad
Since the final velocity (vf) is 0 when the plane comes to a stop, we can rewrite the equation as:
0 = (83.33 m/s)^2 + 2(-5.0 m/s^2)d
Simplifying:
0 = 6944.44 m^2/s^2 - 10.0 m/s^2d
Rearranging the equation:
10.0 m/s^2d = 6944.44 m^2/s^2
Solving for d:
d = 6944.44 m^2/s^2 / 10.0 m/s^2
d = 694.44 m (rounded to two decimal places)
Therefore, the distance it would take for the plane to come to a stop is 694.44 m. Since the actual runway distance is 500 m, the plane will not be able to land safely on this runway.
Given:
Initial velocity, u = 300.0 km/h = (300.0 * 1000) / 3600 m/s = 83.33 m/s
Acceleration, a = -5.0 m/s²
Distance, s = 500 m
Final velocity, v =?
We can use the kinematic equation:
v² = u² + 2as
Substituting the given values:
v² = (83.33)² + 2(-5.0)(500)
v² = 6945.556 + (-5000)
v² = 1945.556
Taking the square root of both sides:
v = √1945.556
v = 44.11 m/s
Since the final velocity of the plane is 44.11 m/s, which is greater than 0, the plane will be able to land safely on the 500 m runway.