[MO412] Question 4
Consider a particle moving in a one dimensional space. It's acceleration is proportional to the square of its velocity. Also, at t=0, it has a velocity v(t=0) = 2 [m/s] and at t=1, v(t=1) = 1 [m/s]. Consider that the particle starts at x=0. A first-order separable ordinary differential equation can be generalized as: dx dt = g ( t ) h ( x ) . With that in mind, choose the alternative that correctly describes the particle's position in relation to time. In other words, x ( t ) , t ≥ 0 . A. x ( t ) = ln ( 1 + t ) 2 B. x ( t ) = 1 + ln ( 2 +
