The position of a certain airplane during takeoff is given by x=1/2 * bt2, where b = 2.0 m/s2 and t = 0 corresponds to the instant at which the airplane's brakes are released at position x = 0. The empty but fueled airplane has an inertia of 35,000 kg, and the 160 people on board have an average inertia of65 kg.

What is the magnitude of the airplane's momentum 15s after the brakes are released?

Question

Shyanne Dougherty

Physics

13 June, 14:15

The position of a certain airplane during takeoff is given by x=1/2 * bt2, where b = 2.0 m/s2 and t = 0 corresponds to the instant at which the airplane's brakes are released at position x = 0. The empty but fueled airplane has an inertia of 35,000 kg, and the 160 people on board have an average inertia of65 kg.

What is the magnitude of the airplane's momentum 15s after the brakes are released?

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Answers (1)

Know the Answer?

Jaden Arnold · Answer 1

1362000 kgm/s

Explanation:

So the total mass combination of the plane and the people inside it is

M = 35000 + 160*65 = 45400 kg

After 15 seconds at an acceleration of 2 m/s2, the plane speed would be

V = 2*15 = 30 m/s

So the magnitude of the plane 15s after brakes are released is

MV = 45400 * 30 = 1362000 kgm/s