A 747 jetliner lands and begins to slow to a stop as it moves along the runway. The mass is 4.6x 10^5 kg, its speed is 27.8 m/s, and the net braking force is 4.42x10^5 N.
(a) What is its speed 7.41 s later?

(b) How far has it traveled in this time?

Question

A 747 jetliner lands and begins to slow to a stop as it moves along the runway. The mass is 4.6x 10^5 kg, its speed is 27.8 m/s, and the net braking force is 4.42x10^5 N.
(a) What is its speed 7.41 s later? 
  
(b) How far has it traveled in this time?

anonymous · Answer

The deceleration can be calculated using 
a = F/m = -0.96 m/s^2

The speed at 7.41s later is 27.8 - (0.96 x 7.41) = 20.7 m/s

Distance travelled = area under speed time graph = area of trapezium = 0.5 x (20.7 + 27.8) x 7.41 = 180m