Starting at t = 0s, a horizontal net force F = (0.275N/s)ti+(-0.460N/s2)t^2j is applied to a box that has an initial momentum p = (-2.90kg?m/s)i+(3.95kg?m/s)j. What is the momentum of the box at t = 2.05s? Enter the x and y components of the momentum separated by a comma.

(-2.32, 2.63) kg m/s

Explanation:

From Newton's second law of motion, force is the rate of change of momentum. Mathematically,

where p is the momentum.

Integrating both sides,

where C is a constant of integration determined by initial values.

The initial momentum is when t = 0.

Hence,

At t = 2.05 s,

Thus, p = (-2.32, 2.63) kg m/s

answer; how far out of place which the object finished relative to the starting location;