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As @AccidentalTaylorExpansion and @David White discuss in their answers, your relationship is only valid for constant linear acceleration. Your situation is rotational, not linear motion. Also, acceleration is a vector and is not constant for circular motion. For circular rotational motion of a particle about a fixed axis with constant magnitude angular ...

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The kinematic equations only apply to situations where the acceleration is constant. In your example the guy on is bike is accelerating at different rates so the kinematic equations do not apply. That's why they refer to $a$ as the acceleration. Edit: after the change of the post the situation now refers to constant acceleration but along a closed path (you ...

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Displacement which you understand, and time which you understand, form the ratio velocity which is displacement divided by time. Consider an object that travels $100 m$ in ten seconds. That means its moved from its starting point to another point $100 m$ away in $10 s$. We then say that it’s velocity is $$v = \frac{100m}{10s} = 10 m/s$$ meaning it travels $... 5 To demonstrate why you can't do this, let's step up one derivative, and explore an anecdote about why it doesn't much sense to add distance and velocity directly. I ask you, "How far away is your house from Jill's house?" You answer, "It is is five kilometers plus seventeen meters per second away." ($5 \,\text{km} + 17\,\text{m/s}$.) I ... 4 The differential equation A general trick in these cases is to solve the differential equation first for the speed, which is easy and first-order, and then integrate the speed to get the position. Indeed, this is a differential equation of the form $$d^2x/dt^2=A-B(dx/dt)^2$$ If we write$v=dx/dt$this becomes $$dv/dt = A-Bv^2$$ which can be solved despite in ... 3 You need an initial velocity. Otherwise, if the object starts at$x=0$then$a=x=0$. Since$v=0$, there will be no motion. 3 Notice that the relative velocity is 2.0m/s. You know the bear was 26m behind the tourist. From this you can find the time the bear will take to catch up with the tourist. This is also the maximum time the tourist has to arrive at his getaway vehicle. Knowing his speed, you can then solve for the maximum distance he was alway from his vehicle. 3 It might be described as pedesis or as random motion. The picture shows the velocities of many particles (as in a gas the molecules all move around). Velocity is a vector: it has a speed and it has direction. In 3 dimensions this means that it has three components mutually at right angles: for present convenience, call them up/down, left/right, towards/away. ... 3 Keep in mind there is a difference between a vector and a component of a vector. The velocity is a vector, the x-component of the velocity (or the "$x$-velocity" in the language your book uses) is just a component of a vector, which is a scalar. The velocity vector can be expanded in terms of unit vectors$\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$(... 3 My interpretation of the question is: is velocity something that really exists in nature or is it just a mathematical construct that we use for calculations? In classical physics the distinction between reality and math is indeed somewhat blurred, and gives a taste of philosophy. However, from the quantum mechanical standpoint the distinction is very "... 3 When we do intuitive physics we use the categories our brains come "prepackaged" with, things like size, duration, force, warmness or colour. As we refine our understanding we start measuring these categories (sometimes splitting them, like colour into intensities at different wavelength). From these more formal variables we can derive other ... 3 What is 'Real'? How do you define 'Reality'? This is not what we deal with in Physics... We just try to get an intuition with the physical quantities while dealing with them. So the question is about whether you can get an intuition for velocity or not. Conservation of an entity is purely incidental and it's just a quality of the universe we live in. 3 The kinematic equations were derived with straight-line motion and constant acceleration in mind. When your hypothetical bicycle rider is riding in a circle, he is never traveling in a straight line. In addition, acceleration is a vector, and in the case of circular motion, acceleration is constantly changing because its direction is constantly changing, ... 3 In physics, rate means rate of change. Basically, how much a certain quantity changes with respect to another quantity that is also changing. You may also sometimes hear the term gradient which describes the same thing. And mathematically, it is also a ratio, say for example $$\frac{\Delta F}{\Delta t}$$ where the numerator is one physical quantity and the ... 2 Light always travels at c in every reference frame - this is one of the core tenets of special relativity. If the lamp is turned on at halfway between A and B, they will both receive the light at the same time as seen by a static observer. There is no way once the light is turned on to travel by vehicle and deliver light faster than the leading edge of the ... 2 Taking the square root of each side means the final velocity equals the initial velocity Not quite -- the final speed (magnitude of velocity) equals the initial speed. It could be$v_f = -v_i$. Your original equation is for 1D motion with constant acceleration (constant in magnitude and direction), so it doesn't directly apply to motion in a circle. However,... 2 The textbook example assumes that the car maintains a constant speed as it goes around the corner, and illustrates that velocity (which is a vector quantity) can change even if speed (which is a scalar quantity) does not change. The car’s average velocity measured over an interval of time before the car turns the corner is$12.5$m/s due south because its ... 2 No, you can't.m/s tells us how fast something is moving/ rate of change of distance and m/s/s tells us the rate of change of speed. we cant subtract them because these two are very different from each other. 2 There is no sensical way to add or subtract them together since they are terms that have completely different units and mean totally different things. Velocity is the rate of change of displacement whilst acceleration is the rate of change of velocity or in other words the rate of change of the rate of change of displacement. If we think of something that is ... 2 You have (by hypothesis) that$u'$and$v$are both$\lt c$and you want to show that$u=\dfrac{u'+v}{1+u'v/c^2} \lt c$. or$\frac{u'}c+\frac vc\lt 1+\frac{u'v}{c^2}$(Clear the denominator and divide both sides of the inequality by$c$). or$\mu+\nu\lt 1+\mu\nu$(Writing$\mu=\frac{u'}c$and$\nu=\frac vc$) or$0\lt 1+\mu\nu-\mu-\nu$or$0\lt (1-\mu)...

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It's about how you define average. The formula $$\bar{v} = \frac{\sum_i v_it_i}{\sum_i t_i}$$ is called a weighted average or a weighted mean, where times $t_i$ are the weights, as opposed to the arithmetic average/mean that is given by the first formula. In some contexts you need to use one, in some contexts the other. You can average over time (then you ...

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We put $y = -h$ and $x = R$ in equation of trajectory. $$-h = R\tan\theta - \frac{gR^2}{2u^2}\sec^2\theta$$ $$-h = R\tan\theta - \frac{gR^2}{2u^2}(1 + \tan^2\theta)$$ Now we have a quadratic equation in $\tan\theta$. As $\theta$ is real the determinant of quadratic equation should be always $\geq 0$. The problem is quite easily solved when you do $b^2 -4ac &... 1 Recall the equation of motion $$v=u+at$$ $$v^2=u^2+2as$$ In your case$a\rightarrow -a$and$v=0$so that $$a=\frac{u}{t} \ \ \& \ \ u^2=2as$$ $$a=\frac{\sqrt{2as}}{t}\Rightarrow \boxed{a=\frac{2s}{t}}$$ Note that you have to put a condition so that when the particle has position$d$its acceleration is set to$0$otherwise the particle gets backward ... 1 Suppose train's and track's clocks are set to zero when the first wagon passes by the bench. The clocks of the wagons are syncronized in the train's frame. In the time$T_0$, according to the bench clock, the clock in the wagon, that is now passing by, is: $$T = T_0\frac{1} {\sqrt{1-\frac{v^2}{c^2}}}$$ As the train has velocity$v$, the length from the first ... 1 The usual definition of a central force is a force such that, with appropriate choice of coordinates, it can be written $$\mathbf F = f(\mathbf r) \hat r$$ where$\hat r$is the unit vector pointing away from the origin. Some authors make a stronger definition, that$f(\mathbf r)$can be written as$f(|\mathbf r|)$, so the force only depends on the distance ... 1 If I understood correctly your notation, we are dealing with and ordinary differential equation here: $$A\ddot{x}=B - C\dot{x}^2$$ If we define function$y=\dot{x}$, it becomes $$A\dot{y} = B -Cy^2,$$ which is trivially solvable $$\frac{Ady}{B - Cy^2}=dt$$ and then integrating both sides. Note that this is a particular case of the Ricatti differential ... 1 It seems OP is essentially asking about the speed of a convex combination $${\bf r} ~=~ \sum_{i=1}^N \alpha_i {\bf r}_i,\qquad \sum_{i=1}^N \alpha_i~=~1, \qquad \alpha_i~\geq~0,$$ in a rigid body. The velocity is then $${\bf v} ~=~ \sum_{i=1}^N \alpha_i {\bf v}_i.$$ Finally the speed is given by the 2-norm$\begin{align} |{\bf v}|~\stackrel{\text{... 1 How about if instead of sound, you're powering your equipment with a water turbine. There, the water does flow in a single direction. It's just a different setup. You're correct, the flow itself is not necessary to transfer power. If instead of a battery we use an AC generator to power the circuit, then the long-term drift of the charges is zero. The ... 1 There is no conservation of velocity because that's just the way this universe works incidentally, we noticed that momentum is conserved in isolated systems and formed a law. Whether or not you can get an intuition for something is purely subjective, your intuition for an entity does not determine its 'existence'. Frankly I find it hard to describe reality, ... 1t$and$t'$are just coordinates and you should treat them the same way as spatial coordinates. In Euclidean geometry you can write any curve parametrically as$x(s), y(s)$, but in special cases you may be able to write$y(x)$or$x(y)$. Likewise you can write a curve in Minkowski space as$x(s), t(s)$or in special cases as$x(t)\$. In point of fact you can ...

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