A truck and a car coming to a stop

[rudransh verma]

Homework Statement

A truck and a car are moving with equal velocity. On applying the brakes both will stop after a certain distance, then assuming both brakes have equal retarding force which one will cover less distance? Or both will cover equal distances?

Relevant Equations

$F=m\frac{dv}{dt}$

If $F$ is constant, then mass is inversely proportional to $dv/dt$. But $dv$ is constant too. So mass is proportional to $dt$. More massive the body more time it will take to stop covering more distance. So car will cover less distance. Is my logic right ?

 

[Lnewqban]

... If retarding $F$ is constant, then mass is inversely proportional to $dv/dt$. [But $dv$ is constant too. So mass is proportional to $dt$.] (this is not correct, I believe) More massive the body more time it will take to stop covering more distance. So car will cover less distance.

Newton's second law.
The more massive vehicle required more force to stop in same distance.
The variation of velocity respect to time was not the same for both vehicles.

 

[jack action]

if $F=m\frac{dv}{dt}$ and you want to get the distance in there, try multiplying both sides by $dx$ and see where it takes you.

The more massive vehicle required more force to stop in same distance.

The question specifies the force is the same and the distance is unknown in both cases.

 

[Steve4Physics]

... then mass is inversely proportional to $dv/dt$.

Yes, but we usually express relationships in this order:
[dependent variable] [type of relationship] [independent variable]

So in this case we would say:
“... then $dv/dt$ is inversely proportional to mass.

But $dv$ is constant too. So mass is proportional to $dt$.

We don't consider $dv$ and $dt$ to be constants or variables. They represent infinitesimally small changes in v and t respectively.

You probably mean $\Delta v$ and $\Delta t$, the overall velocity-change and the overall time.

More massive the body more time it will take to stop covering more distance.

Correct (for the same braking force). But that's only a qualitative description, You might find it a useful exercise to derive formulae for the total time and for the total distance in terms of m, F and u(initial velocity).

 

[kuruman]

You can write the average force as $F=m\frac{\Delta v}{\Delta t}$. Since it is the same for both vehicles, $$m_{car}\frac{\Delta v_{car}}{\Delta t_{car}}=m_{truck}\frac{\Delta v_{truck}}{\Delta t_{truck}}.$$
Can you see what to do (or say) next?

 

[rudransh verma]

You can write the average force as $F=m\frac{\Delta v}{\Delta t}$. Since it is the same for both vehicles, $$m_{car}\frac{\Delta v_{car}}{\Delta t_{car}}=m_{truck}\frac{\Delta v_{truck}}{\Delta t_{truck}}.$$
Can you see what to do (or say) next?

$\frac{m_{car}}{m_{truck}}=\frac{\Delta t_{car}}{\Delta t_{truck}}$
$\Delta t$ is proportional to mass $m$.
I want to ask how to tell which is dependent variable and which is not?

 

[haruspex]

If $F$ is constant, then mass is inversely proportional to $dv/dt$. But $dv$ is constant too. So mass is proportional to $dt$. More massive the body more time it will take to stop covering more distance. So car will cover less distance. Is my logic right ?

Yes, but it would be clearer to write "But $\Delta v$ is constant too. So mass is proportional to $\Delta t$". $\Delta v$ is the whole velocity change over the chosen interval, whereas $dv$ is an infinitesimal.

 

[rudransh verma]

Yes, but it would be clearer to write "But $\Delta v$ is constant too. So mass is proportional to $\Delta t$". $\Delta v$ is the whole velocity change over the chosen interval, whereas $dv$ is an infinitesimal.

Are you saying we cannot use dt or dv when we are dealing with real life senerios because real life events happen in sometime $\Delta t$ not $dt$

 

[haruspex]

Are you saying we cannot use dt or dv when we are dealing with real life senerios because real life events happen in sometime $\Delta t$ not $dt$

I'm saying it is meaningless to say "dv is constant" between the two cases. What is the same in the two cases is the overall change in velocity, $\Delta v$.

 

[rudransh verma]

I'm saying it is meaningless to say "dv is constant" between the two cases. What is the same in the two cases is the overall change in velocity, Δv.

That’s what I wanted to say

 

[Steve4Physics]

$\frac{m_{car}}{m_{truck}}=\frac{\Delta t_{car}}{\Delta t_{truck}}$
$\Delta t$ is proportional to mass $m$.
I want to ask how to tell which is dependent variable and which is not?

Here’s one way to think about it.

If we know the value of a variable before an experiment, we call the variable an independent variable. We often choose what values to use for the independent variable.

But if we don’t know the value of a variable until we perform the experiment, we call the variable a dependent variable.

For the truck/car problem, we know (or can find out) the mass of each vehicle before doing the braking ‘experiment’. So the mass is an independent variable.

But we only find out the values of distance, time and acceleration by doing the experiment. So distance, time and acceleration are dependent variables.

Note that quantities that are held at fixed values are called ‘control variables’. ` For the truck/car problem, the braking force, the initial velocity and the final velocity are the control variables.

 

[rudransh verma]

Correct (for the same braking force). But that's only a qualitative description, You might find it a useful exercise to derive formulae for the total time and for the total distance in terms of m, F and u(initial velocity).

$s=u\Delta t/2$
So car which takes less time to stop covers less distance.

 

[Steve4Physics]

$s=u\Delta t/2$
So car which takes less time to stop covers less distance.

Yes. (But you mean 'vehicle' - you are comparing a truck and a car.)

 

[jack action]

I find it rather confusing to go through the time difference to finally get the distance traveled. As I wrote in post #3, it is much more direct - and less messy - to find the distance directly:
$$F = m\frac{dv}{dt}$$
$$Fdx = m\frac{dv}{dt}dx$$
$$Fdx = m\frac{dx}{dt}dv$$
$$Fdx = mvdv$$
Note that this is only stating that the work $W$ required equals the kinetic energy $E_k$ removed from the vehicle. And we could've started right at that point if we already understood the principle of conservation of energy. If we didn't know, we learned a valid lesson by doing this exercise.
$$dx = \frac{m}{F}vdv$$
$$\int_0^sdx = \frac{m}{F}\int_{v_i}^0vdv$$
$$s = -\frac{mv_i^2}{2F}$$
Assuming $F$ (which is always negative for braking) and $v_i$ are the same in both cases, then:
$$s \propto m$$

 

[kuruman]

find it rather confusing to go through the time difference to finally get the distance traveled.

But the problem is only asking which vehicle covers the less distance. The comparison can be made rigorously without finding expressions for the actual distances: Make a v vs. t plot stating at v₀ and ending at zero for each vehicle. The vehicle whose speed takes less time to drop to zero covers less distance because the area under the curve is smaller. That vehicle can be easily identified because, in this case, the time interval ratio is equal to the mass ratio as OP found in post #6.

 

[Steve4Physics]

The stopping distance (as function of mass) can be very quickly found in a few lines of simple algebra.

a = F/m
v² = u² + 2as
0² = u² + 2(F/m)s
Rearranging gives s in terms of m.

Note s will equal a negative expression because the direction of displacement is opposite to the direction of the force.

 

[jack action]

Make a v vs. t plot stating at v0 and ending at zero for each vehicle.

I have to make a graph? That seems complicated.

 

[kuruman]

I have to make a graph? That seems complicated.

It seemed complicated to me too at first. Then I tried it and I liked it!

 

[jack action]

It seemed complicated to me too at first. Then I tried it and I liked it!

That feels like an "OK Boomer!" moment.

 

[kuruman]

That feels like an "OK Boomer!" moment.

OK.

 

[rudransh verma]

Note that this is only stating that the work W required equals the kinetic energy Ek removed from the vehicle. And we could've started right at that point if we already understood the principle of conservation of energy. If we didn't know, we learned a valid lesson by doing this exercise.

Your proof is very nice.
1.In $Fdx=mvdv$, mvdv is change in kinetic energy. How can you tell? The formula is $\frac12mv^2$
2. What I know about integration is that it is the limit of summation of like $v\Delta t$. So how do you decide when to use this concept as you have here? In short why have you integrated?

 

[jack action]

1.In $Fdx=mvdv$, mvdv is change in kinetic energy. How can you tell? The formula is $\frac12mv^2$

And where does $\frac{1}{2}mv^2$ comes from? It comes from the integration of $mvdv$. See below for more.

2. What I know about integration is that it is the limit of summation of like $v\Delta t$. So how do you decide when to use this concept as you have here? In short why have you integrated?

You have to first understand that $dx$ is not equivalent to $\Delta x$ as pointed out to you in an earlier post:

We don't consider $dv$ and $dt$ to be constants or variables. They represent infinitesimally small changes in v and t respectively.

You probably mean $\Delta v$ and $\Delta t$, the overall velocity-change and the overall time.

Once you solve the integral, $dx$ might become $\Delta x$, but you shouldn't assume that. The case of $dE = mvdv$ is a good example of this. If you want to convert this to the whole speed range quickly you would get:
$$\Delta E = mv_{avg}\Delta v$$
Assuming linear variation, the average velocity $v_{avg}$ would be $\frac{v_f+v_i}{2}$ and velocity difference $\Delta v = v_f-v_i$. $v_f$ and $v_i$ are the final and initial velocity in that range. The energy difference is also $\Delta E = E_f-E_i$.
Putting it together:
$$E_f-E_i =m\frac{v_f+v_i}{2}(v_f-v_i)$$
$$E_f-E_i=\frac{1}{2}m(v_f^2-v_i^2)$$
This is the complete equation (the same one that you would have got by integrating properly). There is not just one velocity variable or one energy variable. But if you set $v_f=v$, $v_i=0$, $E_f=E_k$ and $E_i=0$, then you get the simplified version:
$$E_k = \frac{1}{2}mv^2$$
The important lesson to remember here is that $dv$ is not the same as $\Delta v$.

And in the present problem, you don't want to know about the infinitesimal increase $dx$ covered during the infinitesimal increase $dv$, but the whole distance $\Delta x$ covered (which is $s$ in my previous post, coming from $x_f - x_i = s - 0$) in the speed range $\Delta v$.

Now imagine that your problem included a mass that varied with distance (for example the truck loses its load as it moves) or that the braking force varied with the square of the speed (like braking with the help of aerodynamic drag). The integrals would have been different and led to a much different $\Delta x$.

 

[SammyS]

In reply to @jack action 's Post (#14), you wrote:

Your proof is very nice.
1.In $Fdx=mvdv$, mvdv is change in kinetic energy. How can you tell? The formula is $\frac12mv^2$
2. What I know about integration is that it is the limit of summation of like $v\Delta t$. So how do you decide when to use this concept as you have here? In short why have you integrated?

jack then replied by invoking an argument centered on integration.

Perhaps you understand differentiation and/or differentials better.

So, you ask how is it that $mv\,dv$ is change in kinetic energy.

The derivative of kinetic energy, with respect to velocity, $v$, is:
$\displaystyle \frac{d}{dv} E_K = \frac{d}{dv} \left(\frac12 m\, v^2 \right) = mv$ .

The corresponding differential is: $\displaystyle d \, E_K = mv \, dv$ .

 

[rudransh verma]

Once you solve the integral, dx might become Δx, but you shouldn't assume that. The case of dE=mvdv is a good example of this. If you want to convert this to the whole speed range quickly you would get:
ΔE=mvavgΔv
Assuming linear variation, the average velocity vavg would be vf+vi2 and velocity difference Δv=vf−vi. vf and vi are the final and initial velocity in that range. The energy difference is also ΔE=Ef−Ei.
Putting it together:
Ef−Ei=mvf+vi2(vf−vi)
Ef−Ei=12m(vf2−vi2)
This is the complete equation (the same one that you would have got by integrating properly). There is not just one velocity variable or one energy variable. But if you set vf=v, vi=0, Ef=Ek and Ei=0, then you get the simplified version:
Ek=12mv2
The important lesson to remember here is that dv is not the same as Δv.

I got a short answer why we do integration that to get something from rate of change of that something.
Now you took $dE=mvdv$ where dE is the small area element under the curve of the graph between mv and v. Right?
When we integrate we get the total area E(v)=1/2 mv^2. So what we just did is that we took the limit of the summation of dE?

 

[jack action]

Now you took $dE=mvdv$ where dE is the small area element under the curve of the graph between mv and v. Right?
When we integrate we get the total area E(v)=1/2 mv^2. So what we just did is that we took the limit of the summation of dE?

Right. Yes.