Aparticle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. its initial position is x0 = 1.6 m at t0 = 0 s . at 3.0 s , what is the particle's position?

The velocity function of the particle is v_{x}=2t^{2} m/s.

Since, velocity is change in position with time thus,

v_{x}=\frac{dx}{dt}

rearranging,

dx=v_{x}dt

Initial position is 1.6 m, let the final position be x,

Integrating both side,

\int_{1.6}^{x}dx=\int_{0}^{3}2t^{2}dt

x-1.6=\frac{2(3)^{3}}{3}-\frac{2(0)^{3}}{3}=18

Thus,

x=18+1.6=19.6 m

Thus, position of particle will be 19.6 m.

answer: strength i guess

explanation: