# Displacement and distance when particle is moving in curved trajectory

• NTesla
In summary, the conversation discusses the equations used to solve a problem in Irodov's book. The speaker mentions using the magnitude of the vector ##\vec{v}## to solve the equations and getting the correct answer for displacement. However, the other speaker points out that the equality ##\|\int\vec{v}dt\|\leq\int\|\vec{v}\|dt## only holds for rectilinear motion and not for curved trajectories. They also mention that the equations in the original post are incorrect as they mix scalar and vector quantities. The correct equations should be ##\int_0^T\vec v.dt=l\hat j+\vec u.T## and ##\int_{0}^{\
NTesla
Homework Statement
When we integrate velocity vector wrt time, we get displacement, and when we integrate magnitude of that velocity vector wrt time, we get distance. I've found this to be alright when a particle is in rectilinear motion. However, if the particle is moving in curved trajectory, is it still valid for calculation of displacement and distance ?
Relevant Equations
##\int \vec{v}dt=Displacement##
##\int \left | \vec{v} \right |dt=Distance##
While solving question 1.13(see the attachment) from Irodov, I was doing this: $$\int_{0}^{\tau}(\vec{v}-ucos\theta) dt=l$$, and $$\int_{0}^{\tau}\vec{v}cos\theta dt=u\tau$$. Solving this gave the answer. However, while solving these 2 equations, I only used the magnitude of ##\vec{v}##, and still got the displacement equal as is in RHS. But, if integrating speed wrt time should give distance, how are these 2 equations working to give correct answers?

#### Attachments

• Irodov113.png
12.7 KB · Views: 211
• 10.jpeg
38 KB · Views: 229
Last edited:
Generally the inequality $$\|\int\vec{v}dt\|\leq\int\|\vec{v}\|dt$$ holds. It might be the case that the equality holds, it depends on the formula of the vector ##\vec{v}(t)##, seems to me that in that problem you got such a ##\vec{v}## that the equality holds.

Delta2 said:
Generally the inequality $$\|\int\vec{v}dt\|\leq\int\|\vec{v}\|dt$$ holds. It might be the case that the equality holds, it depends on the formula of the vector ##\vec{v}(t)##, seems to me that in that problem you got such a ##\vec{v}## that the equality holds.
The equality holds only in case of rectilinear motion. When a particle is moving in curved trajectory, the displacement will be less than the distance.

NTesla said:
The equality holds only in case of rectilinear motion. When a particle is moving in curved trajectory, the displacement will be less than the distance.
Hmm... yes I think you are right.

Can you give more details on how you solved the problem. The equations you give at OP seem to be plagued by typos. You substract a scalar from a vector in the first, and equate a vector with a scalar in the second.

Delta2 said:
Can you give more details on how you solved the problem. The equations you give at OP seem to be plagued by typos. You substract a scalar from a vector in the first, and equate a vector with a scalar in the second.
There are no typos. ##ucos\theta## is always in the direction of ##\vec{v}##.

NTesla said:
There are no typos. ##ucos\theta## is always in the direction of ##\vec{v}##.
Strictly speaking ##u\cos\theta## is a scalar it doesn't even have a direction. If by that term you mean the "proper" component of the vector ##\vec{u}## that has magnitude ##\|\vec{u}\|\cos\theta## and has the same direction as ##\vec{v}## then ok but you got to write it properly cause as I said ##u\cos\theta## is a scalar.

I really don't understand. Perhaps you meant to write ##\vec{u}\cos\theta##. But then that means that ##\vec{v},\vec{u}## are in the same direction which of course isn't the case.

In any case I need more details on how you solved the problem, I can't understand exactly what you did.

Delta2 said:
Strictly speaking ##u\cos\theta## is a scalar it doesn't even have a direction. If by that term you mean the "proper" component of the vector ##\vec{u}## that has magnitude ##\|\vec{u}\|\cos\theta## and has the same direction as ##\vec{v}## then ok but you got to write it properly cause as I said ##u\cos\theta## is a scalar.

I really don't understand. Perhaps you meant to write ##\vec{u}\cos\theta##. But then that means that ##\vec{v},\vec{u}## are in the same direction which of course isn't the case.

In any case I need more details on how you solved the problem, I can't understand exactly what you did.
I've added a pic in my original question. Please see it. Once the equation was written, it was just simple calculation, nothing fancy.

Delta2 said:
If by that term you mean the "proper" component of the ##\vec{u}## that has magnitude ##∥\vec{u}∥cos⁡θ## and has the same direction as ##\vec{v}## then ok
Yes I did mean to write that.
But the original question still remains.

Delta2
Sorry hell broke loose here... Had to do somethings.

Your second equation looks ok but essentially takes the x-component of each vector so it is natural that $$\|\int\vec{v_x}dt\|=\int\|\vec{v_x}\|dt=\|\int udt\|=\int\|u\|dt=\|u\|\tau$$

Delta2 said:
Sorry hell broke loose here... Had to do somethings.

Your second equation looks ok but essentially takes the x-component of each vector so it is natural that $$\|\int\vec{v_x}dt\|=\int\|\vec{v_x}\|dt=\|\int udt\|=\int\|u\|dt=\|u\|\tau$$
Yes, the second eq. looks ok to me too. Its the first equation that is causing me trouble..

Neither equation is valid as written because they mix scalar and vector additively.
Alternatively, you could write ##\int_0^T\vec v.dt=l\hat j+\vec u.T##.

Last edited:
Delta2
haruspex said:
Neither equation is valid as written because they mix scalar and vector additively.
Interpreting "u cos(θ)" as the component of u in the direction of v, (|v|-|u| cos(θ)) is the rate at which the particle closes in on its target, so the integral would be the distance covered, not l, the initial separation.
For something to integrate to l you need the component of v orthogonal to vector u.

Similarly, the second integral should use |v|cos(θ).
Alternatively, you could write ##\int_0^T\vec v.dt=l\hat j+\vec u.T##.
The equation that you wrote: ##\int_0^T\vec v.dt=l\hat j+\vec u.T## is right. I understand it.
But I couldn't understand why the 2nd equation(in the original post) was wrong. ##\vec{v}cos\theta## is always in the direction of displacement ##u\tau##.
##\therefore## atleast this equation should be right: ##\int_{0}^{\tau}\vec{v}cos\theta dt=u\tau## should be right. Isn't it ?

NTesla said:
##\therefore## atleast this equation should be right: ##\int_{0}^{\tau}\vec{v}cos\theta dt=u\tau## should be right. Isn't it ?
That equation is manifestly wrong. The LHS is vector, the RHS is a scalar. Correct would be: $$\int_{0}^{\tau}v \cos\theta(t) dt=u\tau$$ where ##\theta(t)## is a function of ##t##, and ##v = |\vec v|##.

ChiralSuperfields, Delta2 and NTesla
PeroK said:
That equation is manifestly wrong. The LHS is vector, the RHS is a scalar. Correct would be: $$\int_{0}^{\tau}v \cos\theta(t) dt=u\tau$$ where ##\theta(t)## is a function of ##t##, and ##v = |\vec v|##.
@haruspex had correctly written: ##\int_{0}^{\tau}\vec{v}dt=l\hat{j}+\vec{u}\tau ##
##\int_{0}^{\tau}\vec{v}dt=\int_{0}^{\tau}(vcos\theta\hat{u}+vsin\theta\hat{j})dt##
##=\int_{0}^{\tau}(vcos\theta dt)\hat{u}+\int_{0}^{\tau}(vsin\theta dt)\hat{j}##
##=\vec{u}\tau +\int_{0}^{\tau}(vsin\theta dt)\hat{j}##

Now since there is no equation stating the function of ##\theta## with time, I'm stuck at this.

NTesla said:
Now since there is no equation stating the function of ##\theta## with time, I'm stuck at this.
I imagine you'll have to generate a differential equation here. Possibly a set of equations.

The solution given by Irodov is as follows:

The first equation(in this pic) is what is bothering me. My understanding is the same as haruspex has mentioned in post#11 above, the integral would be the distance covered, not ##l##, the initial separation.

NTesla said:
The solution given by Irodov is as follows:View attachment 276008

The first equation(in this pic) is what is bothering me. My understanding is the same as haruspex has mentioned in post#11 above, the integral would be the distance covered, not ##l##, the initial separation.
You've misinterpreted that integral. The expression ##v - u \cos \alpha## must be the relative speed between A and B. I.e. the rate at which the distance between A and B is decreasing. The integral of which is the change in distance between A and B, which is also ##l##.

That said, I don't immediately see how he gets that expression.

NTesla said:
The solution given by Irodov is as follows:View attachment 276008

The first equation(in this pic) is what is bothering me. My understanding is the same as haruspex has mentioned in post#11 above, the integral would be the distance covered, not ##l##, the initial separation.
If I may chip-in, I believe the first equation in the pic is simply wrong. The distance covered in the 'y' direction in time ##\tau## is determined by A's y-velocity component. This gives:
##\int_{0}^{\tau}(vsin\theta) dt=l##

Steve4Physics said:
If I may chip-in, I believe the first equation in the pic is simply wrong. The distance covered in the 'y' direction in time ##\tau## is determined by A's y-velocity component. This gives:
##\int_{0}^{\tau}(vsin\theta) dt=l##
That equation is definitely right. But see post #17. Irodov's equation is for the change in distance - although I can't see how he gets it.

Steve4Physics
##(\vec{v}-ucos\theta\hat{v})## is the relative speed of point A wrt point B in the direction of ##\vec{v}##. If we integrate this relative speed wrt time, then why shouldn't we get the total distance covered by point A, instead of getting displacement ##l## ?

PeroK
PeroK said:
That equation is definitely right. But see post #17. Irodov's equation is for the change in distance - although I can't see how he gets it.
OK, I'm with you. Maybe the equation comes from consideration of relative velocity. In A's frame, B has an instantaneous relative velocity of ##v - ucos\theta##. That would produce the required integral.

PeroK
Steve4Physics said:
If I may chip-in, I believe the first equation in the pic is simply wrong. The distance covered in the 'y' direction in time ##\tau## is determined by A's y-velocity component. This gives:
##\int_{0}^{\tau}(vsin\theta) dt=l##
The equation written by you is right. But then, there is no equation stating ##\theta## as a function of time ##t##. So how could this be solved ?

This is one of the problem I have with this question. Whomever I ask this question to, they readjust their understanding of finding distance/displacement in accordance with the solution provided by Irodov.

PeroK
Okay, I see what's he's done. Let's look at the initial situation. A is moving towards B at speed ##v## and B is moving in a perpendicular direction at speed ##u##. The initial separation is ##(l, 0)##. After a short time ##dt##, we have the new displacement of B from A is ##(l - vdt, udt)## and the new distance of B from A is: $$\sqrt{l^2 + v^2dt^2 - 2lvdt + u^2dt^2}$$ Ignoring higher powers of ##dt## gives $$l - vdt$$ In other words, the distance between A and B is reducing at a rate ##v##.

In general, the relative velocity of A with respect to B is ##(v - u\cos \alpha, u\sin \alpha)## and the same applies. The rate of change of distance is the first component only.

NTesla
NTesla said:
This is one of the problem I have with this question. Whomever I ask this question to, they readjust their understanding of finding distance/displacement in accordance with the solution provided by Irodov.
There's a big difference between readjusting your understanding and trying to understand Irodov's solution.

NTesla said:
This is one of the problem I have with this question. Whomever I ask this question to, they readjust their understanding of finding distance/displacement in accordance with the solution provided by Irodov.
It was post #16 before you posted his solution. This is the problem with keeping stuff up your sleeve that you can see and we can't. Now we see how your confusion between velocity and speed arose. But, without us getting to see the solution you were looking at, it's especially hard to work out why you were confused.

NTesla said:
##(\vec{v}-ucos\theta\hat{v})## is the relative speed of point A wrt point B in the direction of ##\vec{v}##. If we integrate this relative speed wrt time, then why shouldn't we get the total distance covered by point A, instead of getting displacement ##l## ?
Because it's the relative speed. The integral gives the change in distance of B as measured from A. If you were traveling with A, you started a distance ##l## from B and finished zero distance from B. In your frame you are always moving directly towards B. So the distance covered (as measured in your frame) is simply ##l - 0 = l##.

If you wanted the distance actually covered by A, it's simply ##\int_{0}^{\tau}vdt##.

PeroK
@PeroK, @haruspex, @Steve4Physics: I've tried to solve it once again. Please let me know if it is correct(Is the reasoning written in lower half of the first pic correct ?):
##\vec{v_{AG}}## is the velocity of particle A wrt Ground. Similarly others.

Steve4Physics and PeroK
NTesla said:
I've tried to solve it once again. Please let me know if it is correct(Is the reasoning written in lower half of the first pic correct ?):
##\vec{v_{AG}}## is the velocity of particle A wrt Ground. Similarly others.
I've also tried this and get the same answer as you (though I haven’t checked your actual working). I found a relatively short/simple method. For info’, here it is (in less-than-rigorous format).

The x-displacements of A and B must be the same during the total time (##\tau##).
##\int_{0}^{\tau}vcos\theta dt =u\tau##

##\int_{0}^{\tau}cos\theta dt = \frac{u\tau}{v}##

From considering the relative motion seen from A’s frame of reference (see previous posts) we know:
##\int_{0}^{\tau}(v - ucos\theta)dt = l##

##\int_{0}^{\tau}vdt – u\int_{0}^{\tau}cos\theta dt = l##

##v\tau – u\frac{u\tau}{v} = l##

##\tau(v^2 – u^2) = vl##

##\tau = \frac{vl}{v^2 – u^2}##

NTesla, Delta2 and PeroK
First, by remembering that for the vector ##x## representing the distance between the two points$$||x||'=\frac{x\cdot\dot x}{||x||}\text{ (for any vector)}\Leftrightarrow \int_0^\tau||x||'dt=-L=\int_0^\tau\frac{x\cdot\dot x}{||x||}dt$$but this is too cumbersome to workout, I think.
Second, using a nice diagram showing the configuration at two instances separated by a small period of time, and the law of cosines to find ##||x||'##.

$$L^2(t+\Delta t)\approx\left(L(t)-u\Delta t\right)^2+(v\Delta t)^2-2\left(L(t)-u\Delta t\right)(v\Delta t)\cos\varphi$$Keeping in mind that ##\cos\varphi=-\cos\theta##, we can develop the expression, and manipulate it to get ##\frac{\Delta L^2}{\Delta t}=...##, then compute the limit as ##\Delta t\to0##, and find ##2L(t)L'(t)=...\Leftrightarrow L'(t)=...##
Finally, ##\int_0^\tau L'(t)dt=-L(0)=\int_0^\tau...dt##

Delta2
NTesla said:
@PeroK, @haruspex, @Steve4Physics: I've tried to solve it once again. Please let me know if it is correct(Is the reasoning written in lower half of the first pic correct ?):
##\vec{v_{AG}}## is the velocity of particle A wrt Ground. Similarly others.
View attachment 276023
Something doesn't look quite right to me with the above, because ##\hat v## depends on time, hence you can't take it out of the integrals.

Delta2 said:
Something doesn't look quite right to me with the above, because ##\hat v## depends on time, hence you can't take it out of the integrals.
I agree, and I very much doubt the validity of the preceding line, where the integral of the component of ##\vec u## orthogonal to ##\vec v## is taken to be zero. That component starts out as directly East (say), then veers to the SE as A swings to the NE. It never has a westward component, so cannot integrate to zero.

Delta2
NTesla said:
The equation that you wrote: ##\int_0^T\vec v.dt=l\hat j+\vec u.T## is right. I understand it.
But I couldn't understand why the 2nd equation(in the original post) was wrong. ##\vec{v}cos\theta## is always in the direction of displacement ##u\tau##.
##\therefore## atleast this equation should be right: ##\int_{0}^{\tau}\vec{v}cos\theta dt=u\tau## should be right. Isn't it ?
Yes, sorry ... it was late here, should have slept on it. Post corrected.

Delta2 said:
Something doesn't look quite right to me with the above, because ##\hat v## depends on time, hence you can't take it out of the integrals.
##||v||## and ##||u||## are given as constants.

archaic said:
##||v||## and ##||u||## are given as constants.
Yes but the direction of ##\vec{v}## (which is the same as the direction of ##\hat v##) varies with time, don't you think?

• Introductory Physics Homework Help
Replies
2
Views
282
• Introductory Physics Homework Help
Replies
7
Views
372
• Introductory Physics Homework Help
Replies
9
Views
852
• Introductory Physics Homework Help
Replies
6
Views
835
• Introductory Physics Homework Help
Replies
1
Views
505
• Introductory Physics Homework Help
Replies
3
Views
827
• Introductory Physics Homework Help
Replies
13
Views
881
• Introductory Physics Homework Help
Replies
1
Views
257
• Introductory Physics Homework Help
Replies
5
Views
712
• Introductory Physics Homework Help
Replies
8
Views
407