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Suppose that a block sliding with positive velocity $v$ on a horizontal surface experiences a drag force $\displaystyle F(v)=-F_0e^{Kv}$ where $F_0$ and $K$ are positive constants. At time $t=0$, the block is at position $x=0$ with initial positive velocity $v_0$. Find the time at which the velocity goes momentarily to zero, and the distance $x$ the block has travelled by this time.

My attempt:

$$F=ma$$

$$\displaystyle m\frac{dv}{dt}=-F_0e^{Kv}$$

$$\displaystyle\int_{v_0}^v -e^{-Kv}dv=\int_0^t\frac{F_0}{m}dt$$

$$\displaystyle\left[\frac{e^{-Kv}}{K}\right]_{v_0}^v=\frac{F_0}{m}t$$

$$\displaystyle\frac{1}{K}(e^{-Kv}-e^{-Kv_0})=\frac{F_0}{m}t \tag{1}$$

When $v=0$,

$$\displaystyle\frac{1}{K}(1-e^{-Kv_0})=\frac{F_0}{m}t$$

$$\displaystyle t=\frac{m}{KF_0}(1-e^{-Kv_0})$$

Now how do I find $x$? I know that I can replace $v$ by $\displaystyle \frac{dx}{dt}$ in (1), but how would I solve the resulting differential equation?

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  • $\begingroup$ first, solve for $v$; you should obtain an equation of the form $v=f(t)$ for some function $f$. Then integrate both sides with respect to $\mathrm dt$. $\endgroup$ – AccidentalFourierTransform Jan 8 '17 at 16:53
  • $\begingroup$ As AFT wrote. The resulting integral is not hard, This is more question for Mathematics. VTC. $\endgroup$ – Gert Jan 8 '17 at 17:22
  • $\begingroup$ The solution of differential equations is mathematics, and can be looked up on the internet or in a mathematics textbook. $\endgroup$ – sammy gerbil Jan 8 '17 at 23:11