Final answer:
The jet plane, with an initial speed of 110 m/s and deceleration of -5.0 m/s², requires 1.21 km to come to a stop. Therefore, it cannot land safely on a 0.8 km runway.
Step-by-step explanation:
A student has asked if a jet plane landing with a speed of 110 m/s and able to accelerate uniformly at a maximum rate of -5.0 m/s2 can stop on a runway that is 0.8 kilometers long. To solve this, we can use the kinematic equation that relates initial velocity, acceleration, and the distance traveled:
d = vit + (1/2)at2
We have the initial velocity (vi) of 110 m/s, and we need to find the time (t) it takes for the plane to stop (final velocity, vf = 0 m/s). Since the acceleration (a) is -5.0 m/s2, we can rearrange the equation to solve for time:
t = (vf - vi)/a
By substituting the values, we find that the time required to stop is:
t = (0 - 110)/(-5) = 22 seconds
Now we can calculate the distance the plane will travel during this time:
d = (110)(22) + (1/2)(-5)(22)2
After calculation, this gives us a distance of:
d = 1210 m, or 1.21 km
Since 1.21 km is greater than the available 0.8 km runway, the plane cannot safely land on that runway.