A plane landing on a small tropical island has
just 50 m of runway on which to stop.
If its initial speed is 47 m/s, what is the
maximum acceleration of the plane during
landing, assuming it to be constant?
Answer in units of m/s
2
.
Vi = 47 m/s
d = 50
===============
v = Vi + a t
0 = 47 + a t
a t = -47
or
a = -47/t
===============
d = Vi t + .5 a t^2
50 = 47 t + .5 a t^2
50 = 47 t +.5(-47/t)t^2
50 = 47 t - .5*47 t
50 = .5*47 t
t = 100/47
then
a = -47/(100/47)= -22 m/s^2
> twice gravity
I think maybe 500 meters, not 50
The method above is correct but
The 0.5 is magically gone on the t equation
The t should of been t = 100/23.5
To find the maximum acceleration of the plane during landing, we can use the equation:
Final velocity squared = Initial velocity squared + 2 * acceleration * distance
Here, the final velocity will be zero since the plane comes to a stop. The initial velocity is given as 47 m/s, and the distance is 50 m. Let's substitute these values into the equation:
0^2 = (47 m/s)^2 + 2 * acceleration * 50 m
Rearranging the equation, we get:
2 * acceleration * 50 m = -(47 m/s)^2
Now, let's solve for the acceleration:
acceleration = (-(47 m/s)^2) / (2 * 50 m)
Simplifying further, we have:
acceleration = -110.63 m/s^2
Therefore, the maximum acceleration of the plane during landing is -110.63 m/s^2 (negative because the plane is decelerating).
To find the maximum acceleration of the plane during landing, we can use the formula for acceleration:
acceleration = (final velocity - initial velocity) / time
In this case, we want to find the maximum acceleration, so the final velocity is 0 m/s (since the plane needs to stop on the runway). The initial velocity is 47 m/s.
Since the time isn't given, we need to find it using the formula for distance traveled during constant acceleration:
distance = (initial velocity + final velocity) / 2 * time
The distance is 50 m and the final velocity is 0 m/s. Substituting these values into the formula, we have:
50 = (47 + 0) / 2 * time
100 = 47 * time
time = 100 / 47 ≈ 2.1277 s
Now that we have the time, we can calculate the maximum acceleration:
acceleration = (final velocity - initial velocity) / time
acceleration = (0 - 47) / 2.1277
acceleration ≈ -22.0856 m/s^2 (negative since it represents deceleration)
Therefore, the maximum acceleration of the plane during landing, assuming it to be constant, is approximately -22.0856 m/s^2.