Max Speed of a Car: A Math Equation Analysis

[maxim07]

Summary:: Question: a car of mass 800kg, 3600N driving force experiences resistive forces of 120v Accelerates from rest. Form equation and show max speed is 30 ms^-1, assuming driving force does not change.

my attempt at a solution -I can not see how to rectify it as I don’t think I did anything wrong in the first steps. Thanks.

 

[BvU]

Does this work when t = 0 ?

$\$

 

[pasmith]

Start with m\dot v = F - kv. Do not subtitute values for m, k and F until the final step.

This is of the form \dot v + f(t)v = g(t) which is easiest solved using an integrating factor, which in this case avoids obtaining \ln |v - 30| = ... and having to work out whether v - 30 is positive or negative.

 

[kuruman]

You don't need to solve the diff. eq. to answer the question.

What is the right-hand side of this equation when the car reaches max speed?

 

[kuruman]

Start with m\dot v = F - kv. Do not subtitute values for m, k and F until the final step.

This is of the form \dot v + f(t)v = g(t) which is easiest solved using an integrating factor, which in this case avoids obtaining \ln |v - 30| = ... and having to work out whether v - 30 is positive or negative.

Integrating factors are your friend if you consider integrals as expressions describing an actual physical situation rather than a mathematical abstraction. In this particular case, we have to do two additions using dummy variables $u$ and $\tau$ to express velocity and time parameters:
1. $\dfrac{du}{u-30}$ from $u = 0$ to $u=v$ (forgive the mixed symbols and numbers).
2. $-0.15~d\tau$ from $\tau = 0$ to $\tau=t.$
Then $$\int_0^v \frac{du}{u-30}=\int_0^t(-0.15)~d\tau~$$takes care of the integration constants automatically. BTW, I cannot think that the absolute value of the argument of a logarithm would ever be needed in an equation describing a physical situation.

 

[pasmith]

Then $$\int_0^v \frac{du}{u-30}=\int_0^t(-0.15)~d\tau~$$takes care of the integration constants automatically. BTW, I cannot think that the absolute value of the argument of a logarithm would ever be needed in an equation describing a physical situation.

What do you think the result of the integral on the left is?

 

[SammyS]

Let's hear from OP .

 

[kuruman]

Let's hear from OP .

I think @pasmith directed the query to me, not the OP.

What do you think the result of the integral on the left is?

Here is my answer: I copied the integral directly from OP's work exactly to make my point. The result of the integral is nonsense because the integrand is clearly negative which it cannot be because the car starts from rest and therefore has to move in the (positive) direction of the force.

Here is what I would do. As you suggested, I would write the equations in symbolic form to end up with $$\frac{dv}{F-kv}=\frac{1}{m}dt\implies \int_0^v \frac{du}{F-ku}=\frac{1}{m}\int_0^td\tau$$"Well", you might ask, "how do I know that the same problem will not arise in this symbolic form of the integrals?"
Answer: If you express the integration variable $u$ as a fraction of the terminal (maximum) velocity $v_{\text{ter.}}$, it should be obvious that nothing horrible happens at any finite time $t$ and that as $t\rightarrow\infty$ the car reaches terminal velocity.

 

[vela]

Here is my answer: I copied the integral directly from OP's work exactly to make my point. The result of the integral is nonsense because the integrand is clearly negative which it cannot be because the car starts from rest and therefore has to move in the (positive) direction of the force.

Why can't the integrand be negative? Since the car starts from rest and accelerates in the $+x$ direction, we would expect $u-30<0$ and $du>0$.

 

[kuruman]

Why can't the integrand be negative? Since the car starts from rest and accelerates in the $+x$ direction, we would expect $u-30<0$ and $du>0$.

I did say that the integrand "is clearly negative" but what followed could have been made clearer. I meant to say that the integrand cannot be negative because the physics says that it must be positive as explained in the rest of the sentence. Because, as you have correctly noted, $u-30<0$ and $du>0$ which makes the integrand mathematically negative, we have a conflict between what the math says and what the physics says. To resolve the conflict, one must accept that this mathematical description, which supposedly applies to this physical situation but results in a negative integrand, cannot be correct.

 

[vela]

I did say that the integrand "is clearly negative" but what followed could have been made clearer. I meant to say that the integrand cannot be negative because the physics says that it must be positive as explained in the rest of the sentence. Because, as you have correctly noted, $u-30<0$ and $du>0$ which makes the integrand mathematically negative, we have a conflict between what the math says and what the physics says. To resolve the conflict, one must accept that this mathematical description, which supposedly applies to this physical situation but results in a negative integrand, cannot be correct.

I still don’t see why you’re saying the physics forbids the integrand from being negative.

 

[kuruman]

I still don’t see why you’re saying the physics forbids the integrand from being negative.

1. At $t=0$, the acceleration is at its maximum value $a=\dfrac{F}{m}$ and the velocity is zero.
2. Because the car starts from rest, it moves in the direction of the force, assumed to be the positive direction. Hence, both $v$ and $dv$ are positive.
3. Once the car starts moving, its acceleration is decreasing because $a=\dfrac{F-kv}{m}$. Nevertheless, the acceleration remains positive. This means that $v$ and $dv$ also remain positive.
4. However, the acceleration cannot decrease forever. When $v$ grows to a value such that $kv=F$, the acceleration becomes zero. Zero acceleration means that the velocity is no longer changing. This means that $dv=0$ and the velocity has reached its terminal value, $v=v_{\text{ter.}}$ and can longer increase.
5. On the basis of statements 1-4, the integrand $\dfrac{du}{F-ku}$ is never negative for $0<u<v_{\text{ter.}}$ because neither the numerator nor the denominator is negative in that interval.

 

[vela]

But that's not the integrand @maxim07 derived, which is what you seemed to be referring to in post 8. For the same reasons you argued that your integrand should be positive, the OP's integrand should be negative. In other words, if the OP's integrand is nonsense, then so is yours.

 

[maxim07]

This is the original question

this is the type of method I am supposed to use for this type of question in my course

answer this is the answer they get

 

[Delta2]

I believe some people here are overcomplicating things. @maxim07 what you did is almost correct, you just should have put $v-30$ into absolute value from the fifth line and after. So your general solution should be $$|v-30|=e^{-0.15t}e^C$$.

You can apply the initial condition now and you ll get $|0-30|=e^C$ or $e^C=30$ which is the correct value for the constant term.

To be able to remove correctly the absolute value so you can recover $v(t)=...$ you have to take cases v>30 or v<30, and again rely on the initial condition to decide on which case the problem setup puts you.

 

[maxim07]

Okay, so considering v<30 -you add 30 to both sides which gives v = 30(1 - e^-0.15t)

 

[Delta2]

if v<30 then $|v-30|=30-v$

 

[vela]

I believe some people here are overcomplicating things.

Well, it wasn't so much overcomplicating things rather than a few of us going off on tangents.

 

[kuruman]

Well, it wasn't so much overcomplicating things rather than a few of us going off on tangents.

I admit going off on a tangent but I wanted to make a simple point. You see, my point was $\dots$ never mind.