Answer:
7 m/s^2
Explanation:
Given that the jet is traveling 37.6 m/s when the pilot receives the message.
And it takes the pilot 5.37 s to bring the plane to a halt.
Acceleration of the plane can be calculated by using first equation of motion
V = U - at
Since the plane is going to stop, the final velocity V = zero.
And the acceleration will be negative
Substitute all the parameters into the formula
0 = 37.6 - 5.37a
5.37a = 37.6
Make a the subject of formula
a = 37.6 / 5.37
a = 7.0 m/s^2
Therefore, the acceleration of the plane to bring the plane to a halt is 7 m/s^2
Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;
At constant speed, we will have;
a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
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The distance is 30 km and the displacement is 22.4 km North East
