How to calculate distance traveled if you know a(v) graph equation?

[malakuka3]

Homework Statement

On a boat going 4m/s the engine stops working. How far does the boat go in the next 10s if the equation for acceleration is a=-k*(v)^2 and k=0.65m^-1? What is the speed of the boat at 10s?

Relevant Equations

v=4m/s
a=-k*(v)^2
k=0.65m^-1

I know that you can find the distance traveled by integrating v(t), but I can't find a way to convert a(v) into it. I tried derivating and integrating the a(v) equation, but I don't know what the results mean.

 

[PeroK]

Please post what you have tried.

 

[malakuka3]

I have tried derivating a(v) and got (-13*v)/10. I have also integrated a(v) and got (-13*(v^3))/30. The problem is I don't know what these results mean.

 

[PeroK]

I have tried derivating a(v) and got (-13*v)/10. I have also integrated a(v) and got (-13*(v^3))/30. The problem is I don't know what these results mean.

I have no idea what they mean either. The first step is to write down the equation of motion, which is:
$$\frac{dv}{dt} = -kv^2$$Do you see that?

 

[malakuka3]

I have no idea what they mean either. The first step is to write down the equation of motion, which is:
$$\frac{dv}{dt} = -kv^2$$Do you see that?

Yes i see that. Does that mean that i can integrate -kv² and get v(t)= (-kv³/3)+t

 

[Orodruin]

Yes i see that. Does that mean that i can integrate -kv² and get v(t)= (-kv³/3)+t

No, you cannot integrate like that. The LHS has a derivative with respect to t, not with respect to v.

You need to solve the differential equation, which in this case is separable.

 

[PeroK]

Yes i see that. Does that mean that i can integrate -kv² and get v(t)= (-kv³/3)+t

No. You must pay attention to the variable with which you are integrating. The next step is
$$-\frac 1 {v^2} dv = k dt$$

 

[malakuka3]

No. You must pay attention to the variable with which you are integrating. The next step is
$$-\frac 1 {v^2} dv = k dt$$

So if I understand this correctly you have to integrate both sides and then you get $$\frac{1}{v} = kt+c$$. After that, you can substitute c for 4, because the velocity at the start is 4m/s. Finally, you can get v out and you are left with $$v=\frac{1}{kt+4}$$

 

[PeroK]

So if I understand this correctly you have to integrate both sides and then you get $$\frac{1}{v} = kt+c$$. After that, you can substitute c for 4, because the velocity at the start is 4m/s. Finally, you can get v out and you are left with $$v=\frac{1}{kt+4}$$

That substitution for $c$ cannot be right. Your final equation is not dimensionally correct.

Also, you must always show the units.

 

[PeroK]

Ps you are better using $v_0$ instead of $4 \ m/s$.

 

[Orodruin]

Also, even forgetting the units, the expression evaluates to 1/4 at t=0.

 

[malakuka3]

Also, even forgetting the units, the expression evaluates to 1/4 at t=0.

Why does c equal to 0.25 m/s (if m/s)?

 

[Orodruin]

Why does c equal to 0.25 m/s (if m/s)?

It doesn’t. Check your units.

 

[malakuka3]

That substitution for $c$ cannot be right. Your final equation is not dimensionally correct.

Also, you must always show the units.

If I understand this correctly, the units for $c$ must be $s/m$ since on the LHS of the equation you have velocity ($m/s$) and on the other, you must also have the same units. So, when you multiply $k$ with $t$ you get $s/m$ and the fraction turns them around, so you get $m/s$. Because of this $c$ has to have $s/m$ as units. Am I assuming correctly?

 

[Orodruin]

If I understand this correctly, the units for $c$ must be $s/m$ since on the LHS of the equation you have velocity ($m/s$) and on the other, you must also have the same units. So, when you multiply $k$ with $t$ you get $s/m$ and the fraction turns them around, so you get $m/s$. Because of this $c$ has to have $s/m$ as units. Am I assuming correctly?

To paraphrase a small green movie hero:
Assume not. Do, or do not. There is no assume.
But yes, everything in your equations must be dimensionally consistent. You can also check that the kt term is dimensionally consistent with your result by figuring out the dimensions of k based on the acceleration equation itself a = -kv^2.

 

[PeroK]

If I understand this correctly, the units for $c$ must be $s/m$ since on the LHS of the equation you have velocity ($m/s$) and on the other, you must also have the same units. So, when you multiply $k$ with $t$ you get $s/m$ and the fraction turns them around, so you get $m/s$. Because of this $c$ has to have $s/m$ as units. Am I assuming correctly?

This is probably a good time in your physics development to move on from a "plug in the numbers" approach. Starting from:

$$\frac{1}{v} = kt+c$$.

At $t = 0$ you have $v = v_0$ and:
$$\frac 1 {v_0} = c$$And that should be clear.

 

[malakuka3]

This is probably a good time in your physics development to move on from a "plug in the numbers" approach. Starting from:

At $t = 0$ you have $v = v_0$ and:
$$\frac 1 {v_0} = c$$And that should be clear.

I see. Now that you wrote it out, the solution for $c$ seems kinda obvious haha. Thanks for the help, I really appreciate it!

 

[malakuka3]

To paraphrase a small green movie hero:
Assume not. Do, or do not. There is no assume.
But yes, everything in your equations must be dimensionally consistent. You can also check that the kt term is dimensionally consistent with your result by figuring out the dimensions of k based on the acceleration equation itself a = -kv^2.

I finally get it now. Thanks so much!