A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.50x^5 kg. Its speed is 27.0 m/s, and the net braking force is 4.30x10^5 N. (a)what is its speed 7.50 s later? (b)how far has it traveled in this time?

Answer 1

Use the braking force and F = ma to get the deceleration rate, a.

(a) After t= 7.5 seconds, the velocity will be
V(7.5) = Vo - a t
= 27.0 m/s - a*7.5

(b) Distance travelled at t = 7.5s
= (Average velocity)* time
= (1/2)*(Vo + V(7.5)) * 7.5s

Answer 2

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