# Find the point when velocity is momentarily zero

• vbrasic
In summary, we have a block traveling in the positive x-direction with some initial velocity ##v_0##. The block is subject to a drag force ##F(v)=-F_0e^{Kv}##, where ##F_0,\,K## are positive constants. We have found the point in time when the velocity of the block momentarily goes to zero to be ##t=\frac{m}{F_0K}(1-e^{-Kv_0})##. The position of the block at this time is ##x=-\frac{mv_0}{F_0K}e^{-Kv_0}-\frac{m}{F_0
vbrasic

## 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. 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 $$\frac{f(x)}{g(x)}.$$

Last edited:
vbrasic said:

## 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.

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

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

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##.

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

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

## 1. What does it mean for velocity to be momentarily zero?

When velocity is momentarily zero, it means that the object is not moving at that particular instant in time. However, this does not necessarily mean that the object has stopped moving entirely.

## 2. Why is it important to find the point when velocity is momentarily zero?

Finding the point when velocity is momentarily zero can help us understand the motion and behavior of an object. It can also provide valuable information about the acceleration and direction of the object at that specific point.

## 3. How do you find the point when velocity is momentarily zero?

To find the point when velocity is momentarily zero, you can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Set v = 0 and solve for t to find the time when velocity is momentarily zero.

## 4. Can velocity be momentarily zero more than once during an object's motion?

Yes, it is possible for velocity to be momentarily zero multiple times during an object's motion. This can occur when the object is changing direction or experiencing varying acceleration.

## 5. How does finding the point when velocity is momentarily zero relate to calculus?

Finding the point when velocity is momentarily zero involves taking the derivative of the position function with respect to time. This is a fundamental concept in calculus, where the derivative represents the instantaneous rate of change at a specific point.

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