Find the point when velocity is momentarily zero

[vbrasic]

Homework Statement

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A block travels in the positive x-direction with some velocity $v_0$. It is subject to a drag force $$F(v)=-F_0e^{Kv},$$ where $F_0,\,K$ are positive constants. Find the point in time when the velocity of the block momentarily goes to zero. Find how far the block has traveled at this time. Take intial time to be $t=0$ and initial position to be $x=0$.

Homework Equations

Just $F=ma$, and some kinematics equations.

The Attempt at a Solution

We have that $$F=m\frac{dv}{dt}=-F_0e^{Kv}.$$ Rearranging gives $$\frac{dv}{e^{Kv}}=-\frac{F_0}{m}dt\rightarrow e^{-Kv}dv=-\frac{F_0}{m}dt.$$ We integrate both sides subject to intial conditions as follows: $$\int_{v_0}^{v}e^{-Kv'}dv'=\int_{0}^{t}-\frac{F_0}{m}dt'.$$ This gives, $$-\frac{1}{K}(e^{-Kv}-e^{-Kv_0})=-\frac{F_0}{m}t\rightarrow e^{-Kv}-e^{-Kv_0}=\frac{F_0K}{m}t\rightarrow e^{-Kv}=\frac{F_0K}{m}t+e^{-Kv_0}.$$ Taking the natural logarithm of both sides gives, $$-Kv=\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}\rightarrow v=v_0\ln{(\frac{F_0K}{m}t)}.$$ So evidently the velocity is zero when $$\ln{(\frac{F_0K}{m}t)}\rightarrow t=\frac{m}{F_0K},$$ My problem arises when I try to find the position of the object at this time. We have, $$v=\frac{dx}{dt}=v_0\ln{(\frac{F_0K}{m}t)}.$$ So, $$dx=v_0\ln{(\frac{F_0K}{m}t)}dt.$$ Integrating to initial conditions, we have, $$\int_{0}^{x}dx'=\int_{0}^{t}v_0\ln{(\frac{F_0K}{m}t')}dt'.$$ So, $$x=v_0t(\ln{(\frac{F_0K}{m}t)}-1)\bigg|_{0}^{t}.$$ However, $\ln{0}$ is undefined. So I'm not sure how to proceed. Rather, I'm quite sure I need to use L'Hospital's rule. Just not sure how to get the former term into the form $$\frac{f(x)}{g(x)}.$$

 

[Ray Vickson]

Homework Statement

[/B]
A block travels in the positive x-direction with some velocity $v_0$. It is subject to a drag force $$F(v)=-F_0e^{Kv},$$ where $F_0,\,K$ are positive constants. Find the point in time when the velocity of the block momentarily goes to zero. Find how far the block has traveled at this time. Take intial time to be $t=0$ and initial position to be $x=0$.

Homework Equations

Just $F=ma$, and some kinematics equations.

The Attempt at a Solution

We have that $$F=m\frac{dv}{dt}=-F_0e^{Kv}.$$ Rearranging gives $$\frac{dv}{e^{Kv}}=-\frac{F_0}{m}dt\rightarrow e^{-Kv}dv=-\frac{F_0}{m}dt.$$ We integrate both sides subject to intial conditions as follows: $$\int_{v_0}^{v}e^{-Kv'}dv'=\int_{0}^{t}-\frac{F_0}{m}dt'.$$ This gives, $$-\frac{1}{K}(e^{-Kv}-e^{-Kv_0})=-\frac{F_0}{m}t\rightarrow e^{-Kv}-e^{-Kv_0}=\frac{F_0K}{m}t\rightarrow e^{-Kv}=\frac{F_0K}{m}t+e^{-Kv_0}.$$ Taking the natural logarithm of both sides gives, $$-Kv=\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}\rightarrow v=v_0\ln{(\frac{F_0K}{m}t)}.$$ So evidently the velocity is zero when $$\ln{(\frac{F_0K}{m}t)}\rightarrow t=\frac{m}{F_0K},$$ My problem arises when I try to find the position of the object at this time. We have, $$v=\frac{dx}{dt}=v_0\ln{(\frac{F_0K}{m}t)}.$$ So, $$dx=v_0\ln{(\frac{F_0K}{m}t)}dt.$$ Integrating to initial conditions, we have, $$\int_{0}^{x}dx'=\int_{0}^{t}v_0\ln{(\frac{F_0K}{m}t')}dt'.$$ So, $$x=v_0t(\ln{(\frac{F_0K}{m}t)}-1)\bigg|_{0}^{t}.$$ However, $\ln{0}$ is undefined. So I'm not sure how to proceed.

Your step
$$-Kv=\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}\rightarrow v=v_0\ln{(\frac{F_0K}{m}t)}$$
is false. You just have
$$v = -\frac{1}{K} \ln\left( \frac{F_0 K}{m}t + e^{-K v_0} \right) ,$$
and it does not really simplify any further.

 

[vbrasic]

Your step

$$-Kv=\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}\rightarrow v=v_0\ln{(\frac{F_0K}{m}t)}$$​

is false. You just have

$$v = -\frac{1}{K} \ln\left( \frac{F_0 K}{m}t + e^{-K v_0} \right),$$​

and it does not really simplify any further.

So basically $$\frac{F_0K}{m}t+e^{-Kv_0}=1\rightarrow t=\frac{m}{F_0K}(1-e^{-Kv_0}),$$ when velocity goes to 0?

 

[vbrasic]

As solutions go, how does this look.

We have, $$F=ma=m\frac{dv}{dt}=-F_0e^{Kv}.$$ This can be rearranged to, $$e^{-Kv}dv=-\frac{F_0}{m}dt.$$ Then, we can integrate both sides subject to specified initial conditions. We have, $$\int_{v_0}^{v}e^{-Kv'}dv'=\int_{0}^{t}-\frac{F_0}{m}dt'\rightarrow -\frac{1}{K}(e^{-Kv})\bigg|_{v_0}^{v}=-\frac{F_0}{m}t\rightarrow \frac{1}{K}(e^{-Kv}-e^{-Kv_0})=\frac{F_0}{m}t.$$ We can rearrange to solve for velocity by, $$e^{-Kv}=\frac{F_0K}{m}t+e^{-Kv_0}\rightarrow -Kv=\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}\rightarrow v=-\frac{1}{K}\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}.$$ Then, to find the instant at which $v=0$, we solve for the zeroes of this express. Because $K$ is a constant, it remains that, $$\ln{(\frac{F_0K}{m}t+e^{-Kv_0})}=0,$$ such that, $$\frac{F_0K}{m}t+e^{-Kv_0}=1.$$ Solving for $t$ we have, $$t=\frac{m}{F_0K}(1-e^{-Kv_0}),$$ when $v=0$. It remains to find position as a function of time, such that we can find the position of the particle when velocity is $0$. We note that, $$F=m\frac{dx}{dt}\frac{dv}{dx}=mv\frac{dv}{dx}.$$ Then we have that, $$mv\frac{dv}{dx}=-F_0e^{Kv}\rightarrow ve^{-Kv}dv=-\frac{F_0}{m}dx.$$ Again, integrating both sides we have, $$\int_{v_0}^{v}v'e^{-Kv'}dv'=\int_{0}^{x}-\frac{F_0}{m}dx'.$$ Integrating by parts we have, $$-\frac{1}{K}v'e^{-Kv'}\bigg|_{v_0}^{v}+\frac{1}{K}\int_{v_0}^{v}e^{-Kv'}dv'=-\frac{F_0}{m}x.$$ Then, $$-\frac{1}{K}(ve^{-Kv}-v_0e^{-Kv_0})-\frac{1}{K^2}(e^{-Kv'}\bigg|_{v_0}^{v})=-\frac{F_0}{m}x,$$ and, $$-\frac{1}{K}(ve^{-Kv}-v_0e^{-Kv_0})-\frac{1}{K^2}(e^{-Kv}-e^{-Kv_0})=-\frac{F_0}{m}x.$$ It remains to find $x$ when $v=0$. We have (when $v=0$), $$\frac{v_0}{K}e^{-Kv_0}+\frac{1}{K^2}e^{-Kv_0}-\frac{1}{K^2}=-\frac{F_0}{m}x.$$ Rearranging gives, $$x=-\frac{mv_0}{F_0K}e^{-Kv_0}-\frac{m}{F_0K^2}e^{-Kv_0}+\frac{m}{F_0K^2},$$ when $v=0$.

 

[PeroK]

That looks correct to me, although I would simply have integrated $vdt$ to get $x$.

 

[PeroK]

PS I would tidy up the final answer by grouping terms. And, check that you get $x=0$ if $v_0=0$.