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

• malakuka3
In summary: 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...This is definitely a good time to start learning about differentiation and integration.
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.

Please post what you have tried.

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.

malakuka3 said:
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?

PeroK said:
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

malakuka3 said:
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.

malakuka3 said:
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
$$-\frac 1 {v^2} dv = k dt$$

PeroK said:
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 $$v=\frac{1}{kt+4}$$

malakuka3 said:
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.

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

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

Orodruin said:
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)?

malakuka3 said:
Why does c equal to 0.25 m/s (if m/s)?

PeroK said:
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?

malakuka3 said:
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
malakuka3 said:
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:
malakuka3 said:
$$\frac{1}{v} = kt+c$$.
At ##t = 0## you have ##v = v_0## and:
$$\frac 1 {v_0} = c$$And that should be clear.

Lnewqban
PeroK said:
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!

Lnewqban and PeroK
Orodruin said:
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!

## 1. How do I interpret an a(v) graph equation?

To interpret an a(v) graph equation, you need to understand that "a" represents acceleration and "v" represents velocity. The equation will show how the acceleration changes as the velocity changes. This can be represented by a line on the graph, with the slope of the line indicating the acceleration.

## 2. What is the formula for calculating distance traveled from an a(v) graph equation?

The formula for calculating distance traveled from an a(v) graph equation is d = v0t + 1/2at2, where d is the distance, v0 is the initial velocity, a is the acceleration, and t is the time elapsed. This formula is derived from the basic kinematic equations.

## 3. Can I calculate the distance traveled if I only have an a(v) graph?

Yes, you can calculate the distance traveled if you only have an a(v) graph. You can use the area under the curve of the graph to find the distance traveled. This can be done by dividing the graph into smaller sections, calculating the area of each section, and then adding them together to get the total distance.

## 4. How can I use an a(v) graph equation to find the maximum distance traveled?

To find the maximum distance traveled, you can use the formula d = v02/2a. This formula is derived from setting the final velocity to 0 and solving for distance. The result will give you the maximum distance traveled based on the initial velocity and acceleration.

## 5. Is there a simpler way to calculate distance traveled from an a(v) graph equation?

Yes, there is a simpler way to calculate distance traveled from an a(v) graph equation. You can use the average velocity and time elapsed to find the distance traveled. The formula for this is d = vavgt, where vavg is the average velocity. This method is useful when the acceleration is constant.

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