A Zambian Air Force jet lands with a speed of 100m/s and can accelerate at maximum rate of -5m/s as it comes to rest
(a) from the instant the jet touches the runway, what is the minimum time needed before it can come to rest.
(b) can the jet land on the small piece of inland in chilubi where the runway is 0.800km long.state why?
i have a student of electrical electronics
please help this equation to do work
i have a student of electrical electronics
To answer these questions, we need to use the equations of motion.
(a) To find the minimum time needed before the jet comes to rest, we can use the equation of motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, since the jet comes to rest)
u = initial velocity (100 m/s)
a = acceleration (-5 m/s^2)
s = distance (unknown)
Rearranging the equation, we have:
0 = 100^2 + 2(-5)s
Simplifying, we get:
0 = 10000 - 10s
10s = 10000
s = 10000/10
s = 1000 m
Therefore, the minimum distance required for the jet to come to rest is 1000 meters.
(b) The runway in Chilubi is 0.800 km long. To determine if the jet can land on this runway, we need to compare the distance required to stop with the length of the runway.
From part (a), we know that the minimum distance required for the jet to come to rest is 1000 meters.
Since 1 kilometer is equal to 1000 meters, we can convert the length of the runway:
0.800 km = 0.800 x 1000 = 800 meters
Comparing the minimum distance required (1000 meters) to the length of the runway (800 meters), we can see that the runway is shorter than the distance required to stop. Therefore, the jet cannot land on the small piece of inland in Chilubi as the runway is not long enough for it to come to a full stop before reaching the end.