# A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.50*10^5 kg, its speed is 27.0 m/s, and the net braking force is 4.91*10^5 N. What is its speed 7.50...

A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.50*10^5 kg, its speed is 27.0 m/s, and the net braking force is 4.91*10^5 N. What is its speed 7.50 s later and how far has it traveled in this time?

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### 1 Answer

The mass of the 747 jetliner is 3.5*10^5 kg and when it lands its speed is 27 m/s. The application of brakes results in a braking force of 4.91*10^5 N.

The deceleration of the jetliner due to the braking force is equal to 4.91*10^5/3.5*10^5 = `491/350` m/s^2.

As the initial velocity of the jetliner is 27 m/s the deceleration results in a decrease in speed and after 7.5 s it is equal to `27 - 7.5*491/350 ~~ 16.47` m/s

The distance traveled by the jetliner during this period of time is `27*7.5 - (1/2)*(491/350)*7.5^2 ~~ 163.04` m

**After 7.5 seconds the jetliner is moving at 16.47 m/s and the distance traveled by it is 163.04 m**