We know that "a=\\frac{dv}{dt}" , so we have a differential equation to solve :
"\\frac{dv}{dt} = -\\frac{k}{v}"
"vdv = -kdt"
We integrate both sides from "t=0" to "t=t" :
"\\int_{t=0}^{t=t} vdv = -\\int_0^t kdt"
"(\\frac{1}{2}v^2)|^{t=t}_{t=0} = -(kt)|^{t}_0"
"\\frac{1}{2}(v(t)^2-v_0^2)=-kt"
"v(t)^2=v_0^2-2kt"
So we find our final solution :
"v(t)=\\pm \\sqrt{v_0^2-2kt}, 0\\leq t\\leq \\frac{v_0^2}{2k}"
We take "+" or "-" depending on the sign of "v_0" (we take the same sign that the sign of "v_0" ) , as "v(t)" must be continuous. We can also note that this solution exists only for some values of t, as for "\\tau=\\frac{v_0^2}{2k}, v(\\tau)=0" and thus the deceleration is not well defined at "t=\\tau" .
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