Question

A jet plane lands with a speed of 100 m/s and can
accelerate at a maximum rate of 5.00 m/s2 as it comes to
rest. (a) From the instant the plane touches the runway,
what is the minimum time needed before it can come to
rest? (b) Can this plane land on a small tropical island
airport where the runway is 0.800 km long?

Answer 1

Answer:

a) t = 20 [s]

b) Can't land

Explanation:

To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.

a)

$v_{f}=v_{i}-(a*t)$

where:

Vf = final velocity = 0

Vi = initial velocity = 100 [m/s]

a = desacceleration = 5 [m/s^2]

t = time [s]

Note: the negative sign of the equation means that the aircraft slows down as it stops.

0 = 100 - 5*t

5*t = 100

t = 20 [s]

b)

Now we can find the distance using the following kinematics equation.

$x -x_{o}=(v_{o}*t)+\frac{1}{2}*a*t^{2}$

x - xo = distance [m]

x -xo = (0*20) + (0.5*5*20^2)

x - xo =  1000 [m]

1000 [m] = 1 [km]

And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land