# Tag Info

## Hot answers tagged kinematics

36

It is exactly because we have a factor of $\frac 1 2$ in the area formula of a triangle. To understand what I'm saying, consider what is the $v(t)$ graph of a particle under constant acceleration. Some say, a good plot is worth a million words! :)

8

You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...

3

You should realize that the first equation you write gives you the value of $a$. In these kinds of problems, you are always given some force, and you are expected to apply Newton's laws to the problem. Therefore $F_{\mu} = -mg\mu \underbrace{=}_{\text{$2^{nd}$Law}} ma \quad \quad \to \quad \quad a = -g\mu$ Now you know the acceleration, you can find how ...

3

Matrices It's a general property of square matrices (and 2-tensors). Any matrix $M_{ij}$ can be decomposed into a part containing the trace, and a part that is traceless. So we begin $$M_{ij} = \frac{1}{N} (\operatorname{tr} M) \delta_{ij} + \left( M_{ij} - \frac{1}{N} (\operatorname{tr} M) \delta_{ij}\right)$$ I hope that much is evident. The first part ...

3

This question can't be answered because important information is not given. We have no idea how much influence, if any, the wind has on the boat's velocity. In the limiting case, if the boat's wind resistance is 0, then the wind has no effect. The question also asks what the boat would do in still water, but we were never told how still or not the water ...

2

There's a more straightforward calculation. In order to travel eastward, the plane's velocity must have a southward component of 60kph to cancel the wind from the south. Since the plane's speed is 100kph, we have the eastward component (in kph) is just: $$v_E = \sqrt{100^2 - 60^2} = 80$$ Thus, you are correct; the time required to travel 189km eastward ...

2

I believe this decomposition has a specific name "Cauchy–Helmholtz theorem (regarding the decomposition of the velocity of a point within an infinitesimal continuum particle)". You can read about this decomposition in the wikipedia page on Strain rate tensor, sections about symmetric and antisymmetric parts and shear rate and compression rate (though this ...

1

Warning : I absolutely don't know nothing about the subject. If you are a french locutor, applyed notion of kinematical inversion (and dynamical inversion) to the Tottori earthquake are explained in this thesis (Sara Di Carli), see page $4$ for a abstract, and pages $11-29$ for details. The Haskell Model(1964) seems to be the first model. In the ...

1

1 year is approximately $y = 3.15 \times 10^7 s$, and light travels at $c = 3.00\times 10^5 km/s$, so convert the speed of the space probe in km/s into ly/s and then into ly/y. Note that $\frac{km}{s} = \frac{km}{s} \left(\frac{3.15 \times 10^7s}{1 y}\right) \times \left(\frac{ly}{3.00\times 10^5 km/s \times 3.15 \times 10^7s}\right) = \frac{1}{30.0\times ... 1 All correct. Your values for$ \dot{x}_P$and$ \dot{x}_C$are negative because you defined$ \dot{x}_R$to be positive (both boats are moving against the river). No problem there, this is completely up to you, but to satisfy the requirement of the question you are required to state that motion of the river has been designated by you as positive direction ... 1 Changing the "wind" to a "current" in an attempt to save the question, the OP's solution is still not quite correct. By subtracting the current's west component from the boat's overall, west velocity over the ground, the OP has correctly found the west component of the boat's velocity through the water. However, he has forgotten that the boat must cancel ... 1 It's not missing, it's in the$\ddot{R}(t)$matrix. It doesn't show up on its own when you do the calculation with matrices instead just vectors however. Vector equations The first equation should be written $$\mathbf{x}_A(t) = \mathbf{\phi}(t) \times \mathbf{x}_B(t)$$ (just reverse where you have$A$and$B$) since given the position in frame$B\$ you ...

1

You may be overthinking it. In general when you need to solve for coefficients you first need to ask what relates the coefficients? In this case it doesn't seem you've written precisely what equation you want. In your post, you've written the parabola in terms of x. Do you mean you want y position as a function of x position? This isn't the standard way to ...

1

They're not the same thing. They have very different implications. You can imagine Force and thus Momentum as the "push" that will happen to the target, while Kinetic energy is the damage it causes. E(k) is equal the Work the object will perform, let it be penetration, fracture, etc. As soon as the object hits the target, the E(k) applies (i.e. the ...

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