A system of four particles moves along one dimension as follows:

particle 1: m₁ = 1.81 kg, v_1(t) = (5.87 m/s) + (0.211 m/s²) t

particle 2: m2 = 3.17 kg, v_(2t) = (9.15 m/s) + (0.399 m/s²) 1

particle 3: m3 = 4.49 kg, v_3(t)=(7.21 m/s) + (0.447 m/s²) t

particle 4: m4 = 4.91 kg

The center of mass of the system is at rest, and the particles do not interact with any objects outside of the system.
Determine the velocity v4 of particle 4 at 1₁ = 2.99 s.

The location of the center of mass of a system is equal to the weighted average of the particles' positions. Similarly, the speed of the center of mass is equal to the weighted average of the particles' speeds.

v = (m₁ v₁ + m₂ v₂ + m₃ v₃ + m₄ v₄) / (m₁ + m₂ + m₃ + m₄)

At time t = 2.99 s, the velocities of each particle are:

v₁ = 5.87 + 0.211 (2.99) = 6.501 m/s

v₂ = 9.15 + 0.399 (2.99) = 10.34 m/s

v₃ = 7.21 + 0.447 (2.99) = 8.547 m/s

Given that the center of mass is at rest, v = 0. Plugging in values and solving for v₄:

0 = (1.81 (6.501) + 3.17 (10.34) + 4.49 (8.547) + 4.91 v₄) / (1.81 + 3.17 + 4.49 + 4.91)

0 = 1.81 (6.501) + 3.17 (10.34) + 4.49 (8.547) + 4.91 v₄

v₄ = -16.9 m/s