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6

First I just want to point out that saying that the four velocity $u_\mu$ satisfies $u_\mu u^\mu=-1$ is a convention, it is not a requirement. It amounts to a choice of the parameterization $\tau$. However, it is a very useful parameterization, it's not common to use other choices. In this parameterization, the four velocity takes the form \begin{equation} ...


4

The derivatives with respect to $\tau$ very much are numbers, but they are not all $1$. Consider your worldline as a curve $\gamma$ parameterized by $\lambda$. We have \begin{align} \gamma : \mathbb{R} & \to \mathbb{R}^4 \\ \lambda & \mapsto (x^0, x^1, x^2, x^3). \end{align} At any point in your worldline you have a position $(x^0, x^1, x^2, x^3)$, ...


3

In the reference frame of the car, the axle is stationary, but the ground is moving below at speed $v_C$. If the car doesn't skid, then the surface of the tyre must move at the same speed, but with a velocity that is directed backwards in the bottom and forwards in the top. At half the distance between the tyre surface and the hub, the speed is $v_C/2$. ...


2

Why is the scalar product of four-velocity with itself -1 The scalar product is invariant In the coordinate system in which the object is (momentarily) at rest, the only non-zero component is the temporal component. See that, in the rest frame, $\gamma = 1$ thus $d\tau = dt$. Then, (setting $c = 1$) we have $$\frac{dx^0}{dt} = 1,\,\frac{dx^i}{dt}=0 ...


2

It is the velocity with respect to the end of the gun barrel. However, it is probably an imprecise term because if you want it down to the mm/s you would have to subtract the recoil speed of the gun. In practice people just measure the velocity with respect to a closely spaced meter on the bench and assume the gun recoil is negligible, as well as the ...


1

If the initial speed (red arrow) is $\vec{u}=(u_x,u_y)$ then the new speed is: $$ \vec{v} = \vec{u} - (\vec{u}\cdot\hat{l})\hat{l} $$ where $\hat{l}=(-\sin \beta,\cos\beta)$ (I'm assuming that the $\beta$ in your diagram is positive). Plugging this in gives \begin{align} v_x &= u_x + k\sin\beta \\ v_y &= u_y - k\cos\beta \end{align} where $$ ...


1

It basically means that they just need to cover half the distance. So, you have the distance to be covered, initial velocity(40) and final velocity has to be zero. Finding deceleration won't be an issue.


1

When you take the average of two quantities, you need to consider the "weighting". In this case, the time spent at each of the velocities matters, and becomes this "weight". In general, when you have a weighted average you multiply each value by its weight, and divide by the sum of the weights. When all the weights are $1$, that reduces to the familiar ...


1

I suspect your teacher glanced at it very quickly, and didn't realize you were using timesteps. When you work with timesteps you are doing a discrete approximation of the differential equations which describe the relationships, namely: $\frac{dv}{dt} = a$ and $\frac{dx}{dt} = v$ which in the discrete form, and arranged to match your code, become: $\Delta ...



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