A 747 jetliner lands and begins to slow to a stop as it moves along the runway. if its mass is 3.50 x 10^5 kg. it speed is 75.0 m/s and the net breaking force is 7.25 x 10^5 N, what is its speed 10.0 seconds later? how far has it traveled in this time?

The speed after 10 s is 54.29 m /s.The distance traveled at that time is 646 m.

Explanation:

First calculate the deceleration rate:

                                     a = F/m

                                       = (7.25 10^5 N) / (3.5 10^5 kg)

                                    a = 2.071 m/s^2 .

At that deceleration rate, the speed after 10 s later is reduced by,

                 speed = 10 2.071

                             = 20.71 m / s making the speed as 54.29 m/s at that time.

  (i.e. subtracting the two speeds = 75 m/s - 20.71 m/s = 54.29 m / s.)

For the distance traveled, multiply the average speed by 10 s.

          Average speed =( 75 m/s + 54.29 m/s) / 2 = 64.645 m / s.

      distance traveled = average speed time

                                     = 64.645 10

distance traveled = 646 m.

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