[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: $\frac{\mathrm{dx}}{\mathrm{dt}}=g\left(t\right)h\left(x\right)$.

With that in mind, choose the alternative that correctly describes the particle's position in relation to time. In other words, $x\left(t\right), t\ge 0$.

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A. $x\left(t\right)=\genfrac{}{}{0.1ex}{}{\ln (1+t)}{2}$

B. $x\left(t\right)=1+\ln (2+t^2)$

C. $x\left(t\right)=2+\ln (1+t^2)$

D. $x\left(t\right)=2\ln (1+t)$

E. None of the above.

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Original idea by: Vitor Antônio Pimenta Silva