Question

In a maneuver, the jet comes in for a landing on solid ground with a speed of 115 m/s, and its acceleration can have a maximum magnitude of 6.72 m/s^2 as it comes to rest.
(a) From the instant the jet touches the runway, what is the manimum time interval needed before it can come to rest?
(b) can this jet land at a small tropical insland airport where the runway is 0.800km long/ Explain the distance he traveled during this negative acceleration.

Answer 1

Answer:

17.1130952381 s

No

Explanation:

t = Time taken

u = Initial velocity = 115 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6.72 m/s² (negative as it is decelerating)

From the equations of motion

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-115}{-6.72}\\\Rightarrow t=17.1130952381\ s$

The minimum time required to stop is 17.1130952381 s

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-115^2}{2\times -6.72}\\\Rightarrow s=984.00297619\ m$

The distance that is required for the jet to stop is 0.98400297619 km which is greater than 0.8 km. So, the jet cannot land on a small tropical island airport.