A 747 jetliner lands and begins to slow to a stop as it moves along the runway. If its mass is 3,50*10^5kg, its speed is 75.0m/s, and the net braking force is 7.25*10^5N, (a) what is its speed 10.0s later? (b) How far has it traveled in this time?

a)

F = net braking force acting on jetliner = - 7.25 x 10⁵ N (the negative sign indicates that the braking force acts to slow down the motion.)

m = mass of the jetliner = 3.50 x 10⁵ N

a = acceleration of the jetliner

acceleration of the jetliner is given as

a = F/m

a = ( - 7.25 x 10⁵) / (3.50 x 10⁵)

a = - 2.1 m/s²

t = time interval = 10 s

v₀ = initial velocity of jetliner = 75 m/s

v = final velocity after time "t" = ?

Using the kinematics equation

v = v₀ + at

v = 75 + ( - 2.1) (10)

v = 54 m/s

b)

x = distance traveled in time "t"

using the kinematics equation

x = v₀ t + (0.5) at²

inserting the values

x = (75) (10) + (0.5) ( - 2.1) (10) ²

x = 645 m