# A truck and a car coming to a stop

• rudransh verma
In summary: Read moreIn summary, the conversation discusses the relationship between mass, velocity, and time in the context of Newton's second law. It is mentioned that if the retarding force is constant, then mass is inversely proportional to the change in velocity over time. However, it is clarified that the change in velocity, represented by dv, is not constant but rather infinitesimally small. The conversation also touches on the concept of dependent and independent variables, with mass being considered an independent variable and distance, time, and acceleration being dependent variables. Finally, the conversation concludes that the more massive the vehicle, the longer it will take to stop and cover a shorter distance.
rudransh verma
Gold Member
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 ?

rudransh verma said:
... 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.

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.

Lnewqban said:
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.

rudransh verma said:
... 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.

rudransh verma said:
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.

rudransh verma said:
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).

jack action
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?

Lnewqban
kuruman said:
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?

rudransh verma said:
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.

haruspex said:
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##

rudransh verma said:
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 ##.

haruspex said:
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

rudransh verma said:
##\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.

Steve4Physics said:
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.

rudransh verma said:
##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.)

rudransh verma
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:
$$s \propto m$$

rudransh verma and hutchphd
jack action said:
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.

Lnewqban
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.

vcsharp2003, jack action and Lnewqban
kuruman said:
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.

jack action said:
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!

Steve4Physics
kuruman said:
It seemed complicated to me too at first. Then I tried it and I liked it!
That feels like an "OK Boomer!" moment.

jack action said:
That feels like an "OK Boomer!" moment.
OK.

jack action said:
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?

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rudransh verma said:
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.

rudransh verma said:
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:
Steve4Physics said:
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##.

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

rudransh verma said:
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 ## .

jack action
jack action said:
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?

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rudransh verma said:
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.

rudransh verma

## 1. How do the weight and size of a truck and a car affect their stopping distance?

The weight and size of a vehicle can greatly impact its stopping distance. Generally, a larger and heavier vehicle, such as a truck, will require a longer distance to come to a complete stop compared to a smaller and lighter vehicle, such as a car. This is because the larger vehicle has more momentum and requires more force to slow down.

## 2. What factors contribute to a truck and a car's ability to come to a stop?

Several factors contribute to a vehicle's ability to come to a stop, including its speed, weight, size, and braking system. The speed at which a vehicle is traveling greatly affects its stopping distance, as does its weight and size. Additionally, the type and condition of a vehicle's brakes play a crucial role in its ability to come to a stop quickly and safely.

## 3. How do weather and road conditions affect a truck and a car's stopping distance?

Weather and road conditions can significantly impact a vehicle's stopping distance. In wet or icy conditions, the road's surface becomes slick, reducing the friction between the tires and the road. This can increase a vehicle's stopping distance, making it more difficult to come to a stop quickly. Additionally, poor visibility due to fog, rain, or snow can also affect a driver's reaction time and make it harder to stop in time.

## 4. What role does a driver's reaction time play in a truck and a car's stopping distance?

A driver's reaction time is a crucial factor in a vehicle's stopping distance. The time it takes for a driver to perceive a potential hazard and react by applying the brakes can greatly impact how quickly the vehicle comes to a stop. A shorter reaction time can help reduce the vehicle's stopping distance and potentially prevent an accident.

## 5. How can a truck and a car's stopping distance be improved?

There are several ways to improve a vehicle's stopping distance, such as maintaining proper tire pressure, regularly checking and replacing brake pads, and avoiding distractions while driving. Additionally, driving at a safe and appropriate speed can greatly reduce a vehicle's stopping distance and help prevent accidents.

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