A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3.60×10^5 kg , its speed is 26.1 m/s , and the net braking force is 4.30×10^5N, what is its speed 7.79 s later? How far has it traveled in this time?

Question

getwaytosid · Answer

s = ut+at^2/2
s= displacement
u= initial speed
a= accelaration (in this case deaccelaration)
t= time interval
s= 26.1*7.79 - (4.3*7.79*7.79)/3.6*2
s=167.077 m