# Help with question mass/displacement

by djames1009
Tags: displacement, mass
 P: 17 From the initial conditions we know that: $x(0)=0 m$ $v(0)=1 m/s$ I'm going to keep using the notations v(t) and x(t) to try to remind you that v and x are functions of t, which is very important when doing calculus. For a constant mass: $F(v(t))=ma=m\frac{dv(t)}{dt}=-0.5v(t)^2$ So: $\frac{dv(t)}{dt}=\frac{-0.5v(t)^2}{m}=\frac{-v(t)^2}{4}$ Erm... which is a separable ordinary differential equation... this is a lot higher leveled than I thought, is this like a university leveled maths question? You're going to need someone else to confirm the answer since I'm not sure if it was meant to turn out like this lol. Anyway, separating gives: $-\int\frac{1}{v^2}dv=\int\frac{1}{4}dt$ Integrating gives: $\frac{1}{v(t)}=\frac{t}{4}+c=\frac{t+4c}{4}$ $\Rightarrow v(t)=\frac{4}{t+C}$ Since, from the second initial condition: $v(0)=\frac{4}{0+C}=1 m/s$ Then: $C=4$ Giving: $v(t)=\frac{dx(t)}{dt}=\frac{4}{t+4}$ Integrating with respect to t: $\int\frac{dx(t)}{dt}dt=\int\frac{4}{t+4}dt$ Gives: $x(t)=4ln(t+4)+b$ Since, from the first initial condition: $x(0)=4ln(0+4)+b=0$ Then: $b=-4ln(4)$ Which gives the final equation for displacement, x(t): $x(t)=4(ln(t+4)-ln(4))=4ln(\frac{t+4}{4})$ And: $x(4)=4ln(\frac{4+4}{4})=4ln(2) m$
 P: 17 $0.5=\frac{1}{2}$ So: $a=-0.5v(t)^2/m=-\frac{1}{2}\frac{v(t)^2}{m}$
 P: 9 $-\int\frac{1}{v^2}dv=\int\frac{1}{4}dt$ Hi i've just realised at this point the limits should be, for $-\int\frac{1}{v^2}dv$ should be V and 1. and for $-\int\frac{1}{4}dt$ should be t and 0. Could you give me any help here?