An airplane touches down at a speed of 55.0 m/s on a runway that is 600 m long. The plane can decelerate at a uniform rate of 2.50 m/s^2. Will the plane be able to stop before the end of the runway?
Why did the airplane become a stand-up comedian? Because it wanted to land some laughs! But let's get back to your question. To see if the plane will be able to stop before the end of the runway, we need to calculate its stopping distance.
We can use the equation:
Final velocity squared = Initial velocity squared + 2 * acceleration * distance.
Since the plane wants to stop, the final velocity is 0 m/s. The initial velocity is 55.0 m/s, and the acceleration is -2.50 m/s^2 (negative because it's decelerating).
Plugging in the values, we have:
0^2 = 55.0^2 + 2 * (-2.50) * distance.
Simplifying it, we get:
0 = 3025 - 5 * distance.
Rearranging the equation, we find:
5 * distance = 3025.
Dividing both sides by 5, we have:
distance = 605 m.
So the plane will not be able to stop before the end of the runway because its stopping distance is greater than the length of the runway. Guess it needs to work on its braking skills!
To determine if the plane will be able to stop before the end of the runway, we need to calculate the distance it will take for the plane to come to a complete stop.
Given:
Initial speed (v0) = 55.0 m/s
Acceleration (a) = -2.50 m/s^2 (negative because it is decelerating)
Distance (d) = 600 m
Using the equation of motion:
v^2 = v0^2 + 2ad
where:
v = final velocity (0 m/s, since the plane comes to a stop)
v0 = initial velocity (55.0 m/s)
a = acceleration (-2.50 m/s^2)
d = distance (600 m)
Plugging in the values:
0^2 = 55.0^2 + 2(-2.50)(d)
Simplifying the equation:
0 = 3025 - 5d
Rearranging the equation:
5d = 3025
d = 605 m
From the calculation, we can see that the plane will not be able to stop before the end of the runway. It will require a distance of 605 meters to come to a complete stop, but the runway is only 600 meters long. Therefore, the plane will overrun the runway.
To determine whether the plane will be able to stop before the end of the runway, we can use equations of motion and calculate the distance it takes for the plane to come to a stop.
The first equation we can use is:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s in this case, as the plane comes to a stop)
u = initial velocity (55.0 m/s in this case)
a = acceleration (deceleration in this case, which is -2.50 m/s^2)
s = distance
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Plugging in the values:
s = (0^2 - 55.0^2) / (2 * -2.50)
Now we can calculate the distance s:
s = (0 - 3025) / (-5)
s = 605 m
The distance required to stop the plane is 605 meters. Since the runway is only 600 meters long, the plane will not be able to stop before the end of the runway.
after t seconds,
v = 55.0 - 2.5t
So, it takes
55.0/2.5 seconds to stop
In that time, the jet travels
s = 55.0t - 1.25t^2 meters
Plug in your value for t and see whether s <= 600