# 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, its speed is 27.0 m/s, and the net braking force is 4.94 x 10^5 N. (a) What is its...

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, its speed is 27.0 m/s, and the net braking force is 4.94 x 10^5 N. **(a)** What is its speed 7.50 s later? **(b)** How far has it traveled in this time?

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When the jetliner starts breaking its velocity is `27m/s` . Once it stops velocity becomes 0.

Using Newton's second law F = ma for the motion under breaks;

`-4.94xx10^5 = 3.5xx10^5*a`

` a = -1.41`

(-) sign used for the force because it applies to opposite direction of moving. So the jetliner moves at a deceleration of 1.41m/`s^2`

Using `v = u+at` at `t = 7.5`

`v = 27+(-1.41)*7.5`

`= 16.425`

** So the velocity of the jetliner at 7.5 seconds will be 16.425m/s**.

Note:

You can find the travelled distance using `S = ut+1/2*a*t^2`