29

In equation (7) you have the expression $$−P\ln(P−Rv(n+1))+P\ln(P)$$ But since this a difference between two logarithms you can rewrite the expression (remember $\ln a - \ln b = \ln \frac ab$) as $$P\ln\left(\frac{P}{P−Rv(n+1)}\right)$$ Now you have the logarithm of a dimension-less quantity, as it should be.


16

In the real world, a displacement time graph can never be discontinuous. The only not-so-physical meaning that it has is that the body teleported from one place to another such that its displacement changed instantaneously/discontinuously. And as you can see, it's impossible to define the velocity of this teleportation.


16

The reason behind this problem is that you've not yet simplified the final expression. For example, let's suppose you get a term $\ln (f(v))$ in your final indefinite integral, where $f(v)$ has dimensions and isn't dimensionless. This is, as you noted, weird as a logarithm's arguments should always be dimensionless. But, now if you apply the limits, you get ...


14

Many, albeit far from all, "multiplications" in physics are simplified versions of integrals. For example, in your case, saying $\Delta s = v \Delta t$ comes from the definition of velocity $$ \vec{v} = \frac{d\vec{s}}{dt} $$ for which we can easily see this has come from an integral: $$ \vec{s} - \vec{s}_0 = \int_{t_0}^t \vec{v}(t)dt. $$ This equation -- ...


10

$$\mathbf{a}=\frac{d\mathbf{v}}{dt}=\frac{dx}{dt}\frac{\partial\mathbf{v}}{\partial x}+\frac{dy}{dt}\frac{\partial\mathbf{v}}{\partial y}+\frac{dz}{dt}\frac{\partial\mathbf{v}}{\partial z}=(\mathbf{v}\cdot\nabla)\mathbf{v}$$


9

Basically the multiplication in physics is the same as in mathematics. But there is one more important thing you need to keep in mind. Unlike in mathematics, in physical formulas the values have a number and a unit. Therefore, when multiplying two physical values, then you need not only to multiply their numbers. You also need to multiply their units. Let'...


6

I am a bit confused with your notation so I chose my notation. Assume the components of the position vector $\vec{R}=[x_1,x_2,\ldots,x_{n_R}]^T $are function of the generalized coordinates $q_1,q_2,\ldots,q_{n_Q}$ thus: $x_j=x_j(q_i)$ where $j=1,(1),n_R$ and $i=1,(1),n_Q\quad ,n_Q \le n_R$ We want to obtain the velocity vector $\vec{v}=\frac{d\vec{R}}{dt}...


6

@Fakemod has answered your question perfectly. But I'd like to add an extra point. You should also know that you cannot apply calculus if the graph is non differentiable at a point, not necessarily discontinuous. In such cases, the simple explanation is that the equations describing the motion have changed, hence, it won't make sense to use a direct ...


5

Acceleration is defined as $$\vec{a}=\frac{d\vec{v}}{dt}$$ At maximum height $\vec{v}$ is zero but it doesn't means that $\frac{d\vec{v}}{dt}=0$ because at just the next moment i.e. $dt$ the particle gains some velocity i.e. $dv$ in downward direction.


5

To an extent, you are correct in your assumptions. The formula $$\mathbf{\vec a = \vec \alpha \times \vec R \implies a=R \, \alpha }$$ when $\vec \alpha$ and $\vec R$ are perpendicular to each other, which is quite the general case. The aforementioned formula relates the tangential acceleration $\vec a$ of a point particle placed at a distance $R$ from ...


5

A note on the case you're considering: A disk is an extended body. This means it's a collection of points: and must be treated as such. To speak of the displacement/velocity/acceleration of a point on the disk and of the disk itself - the center of mass of the disk - are two distinct analyses. Particles on the disk Particles of the disk all travel in ...


4

If two events are at $(t,x,y,z)$ and $(d+dt,\; x+dx,\; y+dy,\; z+dz)$ in some given inertial frame then the proper time between them is $d \tau$, given by: $$ c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2 $$ which gives $$ d\tau = \sqrt{dt^2 - (dx^2 + dy^2 + dz^2)/c^2} $$ If you are not used to writing things like $dt^2$ and $dx^2$ then don't worry; keep ...


4

So mathematically if velocity is zero there can not be any acceleration That is the wrong conclusion. Acceleration is a "change in velocity" not velocity itself. Let's look at a simple example. Let's say we fire a projectile upwards an initial velocity of $100m/s$ and acceleration due to gravity is $10m/s^2$. That means that the velocity will be decreased ...


3

This can be a difficult concept at first. But notice the following facts: At the top of the trajectory, for an instant, the velocity is zero. An instant later, the velocity is non-zero. The particle is falling downward. That means, the velocity has changed from zero to non-zero. What is acceleration but a change in velocity? If there were no change in ...


3

The velocity is changing only the direction, which is why there exists only a centripetal acceleration. If the velocity would have been changing its magnitude (in other words, if the speed was changing) then you'd see that tangential acceleration would exist. This can also be seen by the relevent formula of a body's acceleration in polar coordinates: $$\...


3

Yes you can. There are five equations that you can quote (derivation below) for motion in a straight line under constant acceleration that relate the quantities: displacement, $s$, initial velocity, $u$, final velocity, $v$, acceleration, $a$, and time, $t$. They often referred to as SUVAT equations for obvious reasons! The equation that applies to your ...


3

Sure. If you only consider the "polar part" of the motion, then for Newton's law in cylindrical coordinates we have $$\mathbf F=m(\ddot r-r\dot\theta^2)\,\hat r+m(r\ddot\theta+2\dot r\dot\theta)\,\hat\theta+m\ddot z\,\hat z$$ Since the helix has a constant radius, the centripetal component of the force just becomes $$F_c=-mr\dot\theta^2=-\frac{mv^2}{r}$$ ...


3

There are two answers to this question, depending on whether you are an engineer or a physicist, as follows. In an idealized (physics) world where a ball of infinite stiffness strikes a surface of infinite stiffness and mass, the curve will be discontinuous, as you suspect. In a messy (engineering) world, bouncing objects are not infinitely stiff, and if ...


3

in motion graphs, (x,y,z) coordinates depend on t polynomially. That is only a special case (but it may be the only thing you have seen, if you have just begun studying projectile motion). The motion of rigid bodies when they collide is a standard situation where the "motion graph" is not differentiable.


3

Physics can be thought of as the subset of mathematics which happens to make correct predictions about the Real World (tm) [at least roughly speaking]. As such in my view it's probably a mistake to read too much into the "physical meaning of multiplication" beyond the mathematical definitions. Of course there are lots of friendly-sounding examples to give ...


3

No, it is not. Suppose, a body is moving at a uniform velocity $v$, now there is no restriction on how much time it wants to remain with that same velocity. And after sometime it can accelerate if there is a net force on it. Now, acceleration means a rate of change in velocity and obviously it will take some time to increase (or decrease) it's velocity. It ...


3

$$I(v)=e^{\int h(v)dv}=e^{-\ln(P-Rv(n+1))}=\frac{1}{P-Rv(n+1)} \tag{6}$$ If $v$ is has non-trivial dimensions, then $\int \frac{1}{v} dv = \ln\left|\frac{v}{D}\right|$, where $D$ is equivalent to $e^{-C}$ in the dimensionless case: $\int \frac{1}{x} dx = \ln\left|x\right| + C = \ln\left|x\right| - \ln e^{-C}$. If $v$ is in $\left.\mathrm{m}\middle/\mathrm{...


2

Consider a case in which a charged point sized particle is moving towards another due to electrostatic fore. Its velocity-time graph will have a non-differentiable point at the point where the the two charges meet-the acceleration will be undefined. So, motion graphs can have points where curve cannot be differentiated. (it is not a sharp edge, but still, ...


2

By no interactions they mean no intermolecular attraction forces (van der Waal forces). In other words the internal energy of an ideal gas is all kinetic and no potential energy. In addition collisions between molecules and with the walls that contain them are considered perfectly elastic, so that both kinetic energy and momentum are conserved. Hope this ...


2

You can extrapolate average speed from a continuous regions separated by discontinuities. Take a look at such function jump discontinuity case : In this case body average speed would be : \begin{align} v &= \frac {dx(t)}{dt} \\&= \frac 12 \left( \sin(t)^\prime + (2t)^\prime \right) \\&=\frac{\cos(t)+2}{2} \end{align}


2

Why is there no tangential acceleration in uniform circular motion? Because, then we wouldn't call it uniform. Circular motion can have both tangential $a_\parallel$ and centripetal $a_\perp$ acceleration. They are then the components of an acceleration vector $\vec a=(a_\parallel,a_\perp)$ pointing at an angle. Those special cases where there is no ...


2

Uniform acceleration means an even, or linear, change in speed $v$. A linear function is a straight line. So, just draw a straight line. The angle of that straight line depends on the value of $a$; the larger a value, the steeper. You can simply count it out: For each unit (each second) out horizontally, you must go up/down the whole value of $a$. And ...


2

Consider yourself in the motor boat when you are moving down your boats speed as well as rivers speed add up cause you are moving in the direction of river 15=(v+u)45/60 Also when you retrace the path your boats velocity v is opposite to river flow so (V-u)t= 6km Meanwhile the barge has only covered 9km in the whole time U(t+45/60) = 9km Solve by ...


2

You just need to apply $\vec{F}=\frac{d}{dt}(\gamma m\vec{v})$. For example suppose the second case you say, that I apply a force in some direction, let's say the x-direction, to a particle with zero initial velocity. You would get $$F_x=m\frac{d}{dt}(\gamma v_x)$$ $$0=\frac{d}{dt}(\gamma mv_y)$$ $$0=\frac{d}{dt}(\gamma mv_z)$$ From the second equation you ...


2

Suppose you walk past the point $x = 0$. At the moment you pass this point, your position is zero, but your velocity is nonzero. But your own logic would say this is impossible: But mathematically we know that velocity is the first derivative of position with respect to time $(v=dx/dt)$. So mathematically if position is zero there can not be any velocity. ...


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