Question
Asked Sep 13, 2019
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A 747 jetliner lands and begins to slow to a stop as it moves along the runway.

If its mass is 3.56×105 kg, its speed is 73.0 m/s, and the net braking force is 4.30×105N, what is its speed 12.0 s later?

How far has it traveled in this time?

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Expert Answer

Step 1

Acceleration is,

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a т 4.30x10 N 3.56x10 kg -1.21m/s

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Step 2

Since, brakes are applied, acceleration is negative. Using the equation of motion,

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v=u+at (73.0m/s)+(1.21m/s?)(12.0 s) =58.5m/s

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Step 3

Distance travel...

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2-u S = 2a .(58.5m/s)(73.0m/s) 2(-1.21m/s2 = 788.88 m 789m

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Physics

Kinematics

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