G-force experienced by a jumper, using a DE

[JessicaHelena]

Homework Statement

The eqation $mv' = Kv^2 - mg$ describes the velocity (negative velocity points downwards of a parachute jumper of mass m subject to gravity g, and wind resistance from the open parachute of $Kv^2$, with K a constant. m= 100 kg, g=10 m/s^2, K=10 kg/m.

Suppose that at $t = 0$, $v(0) = -20 m/s$. Find the time at which the acceleration $v'$ is largest in absolute value. At that moment, what is the g-force experienced by the jumper?

Relevant Equations

$v' = \frac{1}{10}v^2 - g$.

We are given that $v' = \frac{1}{10}v^2 - g$.

I tried using implicit differentiation so that $v'' = \frac{1}{5}vv' = \frac{1}{5}v(\frac{1}{10}v^2-g)$ and set this equal to 0. Hence we have 3 critical points, at $v= 0$, and $v = \pm \sqrt{10g}$.

Calculating $v''(0)=-120$, we know the function is concave, but I'm not really sure where to go from here.

 

[PeroK]

Please fix your Latex. You can use ## as a delimiter.

 

[JessicaHelena]

Please fix your Latex. You can use ## as a delimiter.

I didn't know this forum used another LaTeX formatting—it should be fixed now.

 

[PeroK]

Homework Statement:: The equation $mv' = Kv^2 - mg$ describes the velocity (negative velocity points downwards of a parachute jumper of mass m subject to gravity g, and wind resistance from the open parachute of $Kv^2$, with K a constant. m= 100 kg, g=10 m/s^2, K=10 kg/m.

Suppose that at $t = 0$, $v(0) = -20 m/s$. Find the time at which the acceleration $v'$ is largest in absolute value. At that moment, what is the g-force experienced by the jumper?
Homework Equations:: $v' = \frac{1}{10}v^2 - g$.

We are given that $v' = \frac{1}{10}v^2 - g$.

I tried using implicit differentiation so that $v'' = \frac{1}{5}vv' = \frac{1}{5}v(\frac{1}{10}v^2-g)$ and set this equal to 0. Hence we have 3 critical points, at $v= 0$, and $v = \pm \sqrt{10g}$.

Calculating $v''(0)=-120$, we know the function is concave, but I'm not really sure where to go from here.

The motion never has $v = 0$. Can you draw a graph of the acceleration?

PS What's the significance of $v = -\sqrt{10g}$. Perhaps draw a graph of $v$ first.

 

[kuruman]

I tried using implicit differentiation so that $v'' = \frac{1}{5}vv' = \frac{1}{5}v(\frac{1}{10}v^2-g)$ and set this equal to 0.

How would this help you find the time at which the acceleration is largest? I would find $v(t)$ first, then $a(t)$, then the time.

 

[PeroK]

How would this help you find the time at which the acceleration is largest? I would find $v(t)$ first, then $a(t)$, then the time.

Perhaps it's simpler than that.

 

[kuruman]

Perhaps it's simpler than that.

Sure it is. The title says " ... using DE." That can be interpreted as solving the DE or just looking at it and imagining what happens to the acceleration as $v$ increases from zero to "some" value.

 

[JessicaHelena]

The problem is I don't really know how to find the time at which the acceleration $v'$ is the largest in absolute value. Would it be one of the points where a = v' = 0? Then I did find (earlier), $0$ and $\pm sqrt{10g}$, but which one of those points matter & how could I know that?

@PeroK is there a simpler way? What would it look like?

@kuruman I could find $v(t)$, but isn't $a(t) = v'$, which is already given?
Also, I did manage to get an expression for $v$, but I'm not sure if it was supposed to look so complicated: $v(t) = \sqrt{10g} tanh(c-t)$ where $c = tanh^{-1} (-20/\sqrt{10g})$.

 

[kuruman]

The problem is I don't really know how to find the time at which the acceleration $v'$ is the largest in absolute value. Would it be one of the points where a = v' = 0? Then I did find (earlier), $0$ and $\pm sqrt{10g}$, but which one of those points matter & how could I know that?

@PeroK is there a simpler way? What would it look like?

@kuruman I could find $v(t)$, but isn't $a(t) = v'$, which is already given?
Also, I did manage to get an expression for $v$, but I'm not sure if it was supposed to look so complicated: $v(t) = \sqrt{10g} tanh(c-t)$ where $c = tanh^{-1} (-20/\sqrt{10g})$.

Nice try. The correct expression for $v(t)$ does involve a hyperbolic tangent, but is not quite what you have. With the correct expression you could then take the time derivative to find the acceleration as a function of time and then find the time at which the maximum value occurs the usual way. Or you can plot it and see when the acceleration is maximum. That's the formal way.

The shortcut is to look at the DE $$m\frac{dv}{dt}=Kv^2-mg$$and, without solving the DE, ask yourself what the magnitude of the acceleration could possibly be initially at t = 0 and then a very long time later. Then ask yourself what it could possibly be relative to the two extremes at intermediate times. Hint: What do you think the acceleration is when the parachutist hits the ground?

 

[JessicaHelena]

@kuruman
When you say to "find the time at which the maximum value occurs the usual way", am I right in thinking that I need to differentiate the acceleration function, find the critical points, and find a point where the values to the left of the critical point (on a number line) are positive and the values to the right of the point are negative?

Regarding the shortcut, I think the acceleration as the parachute hits the ground would be (mg - drag), but if the person reached the terminal velocity, a=0. On the other hand, at t=0, a = g (downwards is positive per the problem) since the drag force is not felt yet. Then would that mean it's g/2?

 

[PeroK]

@kuruman
When you say to "find the time at which the maximum value occurs the usual way", am I right in thinking that I need to differentiate the acceleration function, find the critical points, and find a point where the values to the left of the critical point (on a number line) are positive and the values to the right of the point are negative?

Regarding the shortcut, I think the acceleration as the parachute hits the ground would be (mg - drag), but if the person reached the terminal velocity, a=0. On the other hand, at t=0, a = g (downwards is positive per the problem) since the drag force is not felt yet. Then would that mean it's g/2?

I think you are struggling to analyse the motion in this case.

The jumper starts at $v = -20 m/s$. The terminal velocity with the parachute open is $v = -10m/s$. When the parachute is opened at $t =0$ the drag force is greater than gravity. There is, therefore, an immediate upward, positive acceleration.

The positive acceleration reduces the magnitude of the downward velocity. This reduces the drag force.

The downward motion, therefore, is simply a gradual descrease of the magnitude of the velocity until terminal velocity is reached.

Can you finish off the problem from there?

Note, also, that when you analysed the acceleration for the motion, you didn't notice that there were no turning points within the range of velocity values for this problem. The critical points of $v = 0$ and $v = \sqrt{10g}$ are outside the motion entirely. And the third case, $v = -\sqrt{10g}$, represents the terminal velocity for the motion in this problem.

 

[JessicaHelena]

@PeroK
Frankly, I'm sorry, but I'm a bit frustrated myself because I don't seem to get it. Still, from what I understand, since velocity (magnitude-wise) only decreases until the terminal velocity is reached, the greatest acceleration happens at t = 0.

Hence we come back to the equation given: $mv' = Kv^2 - mg$. Plugging in $v(0)$, we have $mv' = 3000$ and hence $v' = 30$.

At this point, I realized I'm not really sure what g-force means even after having tried to figure it out by watching videos etc. Would I get $30/g = 3$ as my g-force?

 

[JessicaHelena]

Or, I found that the g-force can be calculated using the formula N/mg, where N is the normal force.

I don't think we have a normal force here, but assuming $Kv^2 = "N"$ here, $N = m(v' + g)$. Then g-force is given by the expression $\frac{g+a}{g}$. To find $a$ at $t=0$, we do $v' = Kv^2/m - g = 30$. Then the g-force is $\frac{30+10}{10} = 4$. Is this right?

 

[PeroK]

Yes, as the parachutist is free-falling there is no g-force. When the parachute is opened the force experienced by the parachutist is entirely this upward drag force. And, at $t=0$ the g-force is $4g$.

Note that the g-force, technically, is a force per unit mass, i.e. has the units of acceleration.

 

[JessicaHelena]

Note, also, that when you analysed the acceleration for the motion, you didn't notice that there were no turning points within the range of velocity values for this problem. The critical points of v=0v=0v = 0 and v=√10gv=10gv = \sqrt{10g} are outside the motion entirely. And the third case, v=−√10gv=−10gv = -\sqrt{10g}, represents the terminal velocity for the motion in this problem.

@PeroK
Thank you for clarifying that up!
But I wanted to go back to what you said previously about the terminal velocity. How do I know that $0$ and $\sqrt{10g}$ do not matter here?

 

[PeroK]

@PeroK
Thank you for clarifying that up!
But I wanted to go back to what you said previously about the terminal velocity. How do I know that $0$ and $\sqrt{10g}$ do not matter here?

$v = 0$ ... when does the parachutist come to rest? Apart from when they hit the ground?

$v = \sqrt{10g}$ ... when does the parachutist start moving upwards?