What is the change of momentum on a 45,000 kg airplane when it changes its velocity from 242 m/s to 258 m/s? 720,000 kg•m/s

16 kg•m/s

440,000 kg•m/s

2,800 kg•m/s

720,000

Explanation:

change in momentum = m (v2-v1)

= 45,000(258-242)= 720,000

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west