# Tag Info

9

The velocity is constantly increasing due to a constant acceleration. Exactly at 1 s the velocity is 10 m/s, but this does not mean that velocity was at 10 m/s in preceding second. In fact, given the distance 5 m moved in this second, the average velocity in this second was 5 m/s. And this should make sense to you, because in this first second the velocity ...

7

In terms of velocitiy in classical mechanics the only difference between the dimensions is what kind of objects $v$, $u$ and $a$ are. While $t$ should always be a real number $v$,$u$ and $a$ should be vectors in $\mathbb{R}^n$ where $n$ represents the dimension you are talking about. If this does not help, please clarify.

4

Like all paradoxes, there is no contradiction here, just misuse of logic. How do you define velocity? If you say the distance traveled in an extended period of time, divided by that time well then of course there's no such thing as instantaneous velocity. Asking what something's instantaneous velocity is under this definition is logically equivalent ...

4

The equation comes from Newton's second law: $$F = ma$$ Galileo didn't know calculus (because Newton and Leibniz hadn't discovered it yet) so he couldn't derive the equation mathematically. Since we do know calculus we know that acceleration is the variation of velocity with time: $$a = \frac{dv}{dt}$$ And also the gravitational force $F$ is equal to ...

3

Notice that in more than one dimension (two, three, and even higher!) the boldface vector notation $\mathbf x(t)$ is really just shorthand for a tuple of component functions. Explicitly, in dimension $d$, one has \begin{align} \mathbf x(t) = \begin{pmatrix} x^1(t) \\ \vdots \\ x^d(t) \\ ...

3

You use for example python and mathplolib like in the following (slightly modified) example taken from http://faculty1.coloradocollege.edu/~sburns/toolbox/ODE_II.html. You need a working python/pylab installation to run it via python trajectory.py (if you save the file with this name) from the command line. The idea is to solve newtons equation of motion ...

3

Relative position versus time is linear when relative velocity is constant i.e. when relative acceleration eqauls zero. In this case, both stones are acted upon by gravity only and hence their acceleration always equals $g$ directed downwards. Therefore, relative position varies linearly for the entire duration of motion which stops when the stone thrown ...

3

If you assume that Your body is a uniform, thin rigid rod. One end of the rod is pivoted (aka your feet) during the fall. Then one simply recalls that the angular velocity $\omega$ of rotation of your body is related to the tangential velocity $v$ of a point a distance $r$ from the pivot by \begin{align} v = \omega r \end{align} Now, if you have height ...

2

The magnitude of acceleration is simply a measurement of change in speed per unit time. As an example, say you are in a car starting from rest and you begin to speed up. Say that you reach a speed of $20 {m \over s}$ in $2$ seconds. This means the magnitude of your acceleration is: $$a = {20 {m \over s} \over 2s} = 10 {m \over s^2}$$ That is, your speed ...

2

Your question is kind of vague but I will try to respond. Acceleration is defined as the time rate of change of velocity. Since velocity has both magnitude and direction, so does acceleration. In other words, acceleration is a vector. The length of the vector is its magnitude. Its direction is the direction of the vector. So the magnitude of ...

2

You're on the right track. A couple of notes: Those are actually total derivatives. You can think of $x(t)$ and $y(t)$ as functions of $t$ alone you have two equations for two functions. You probably want to isolate them into two equations, each for one function Think about how you would solve this by elimination

2

You did not carry out your integration quite correctly. We have: $$a(t)= -B_0+B_1t$$ $$v(t) = \int\! a(t)dt=-B_0 t+\frac{1}{2}B_1t^2+C$$ Then plugging in conditions to solve for $C$ we get: $$v(t_s)=0$$ $$0=-B_0t_s+\frac{1}{2}B_1t_s^2+C$$ $$C=B_0t_s-\frac{1}{2}B_1t_s^2$$ Now we can plug in $t=0$ and solve for $v(0)$ $$... 1 The value of initial velocity will be different for different angles θ with the horizontal.. So I got this result.$$ u=(gR/(sinθcosθ-cos^{2}θ))^{1/2} $$or$$ u=(2gR/(sin2θ-2cos^{2}θ))^{1/2} $$or$$ u=(39.2/(sin2θ-2cos^{2}θ))^{1/2} $$This is my attempt for the solution(i have attached image): From A to B displacement is FB From C to B displacement is ... 1 I am not sure what you meant by: "I figured I could simply calculate the magnitude of the components since that will give me the distance" But the idea is use the kinematics equations for x and y: x(t)=x_{0}+v_{x0}t+1/2at^2 and y(t)=y_{0}+v_{y0}t+1/2at^2 These equations are derived from integrating the acceleration function ... 1 Suppose you have a satellite of mass m at a distance r. If we assume the satellite is small enough to behave as a point mass the moment of inertia of the satellite is:$$ I = m r^2 $$so its kinetic energy is:$$ E = \tfrac{1}{2} I w^2 = \tfrac{1}{2} m r^2 \omega^2 \tag{1}$$But for a body moving in a circle of radius r at an angular velocity ... 1 The canoe has a relative velocity to the river which we simply call \vec{v}_{c/r} =( \dot{x}, \dot{y} ). The motion of the river relative to the earth is \vec{v}_{r/e} = ( 0.54, 0 ). The canoe relative to the earth is \vec{v}_{c/e} = ( 0.55\cos(-45^\circ), 0.55\sin(-45^\circ) ) . All together you have$$ \vec{v}_{c/e} = \vec{v}_{c/r} + \vec{v}_{r/e} ...

1

This is correct, assuming constant acceleration, we have for this problem $$a = -\frac{v^2_i}{2d} = -\frac{(6.26)^2}{0.05} = -784\frac{m}{s^2}$$ First, I applaud you for asking the question. Too often, I have graded homework and tests where numbers were submitted for answers without any thought as to whether they were reasonable. However, this is a ...

1

Well, let's think about this: An object is traveling at 6.26 m/s during impact and travels only 0.025 meters before stopping. The force causing the object to decelerate needs to be extremely high. It's just like having a force being applied for a very short period of time, such as a bat hitting a baseball and the time of contact is extremely small, you ...

1

Let's take a look just from the point of view of someone reading the problem. First of all, we can say "our car has motion", because it's changing its position each second. Ok. So, how is its motion? Well, it is moving in 1 dimention, it is a linear movement. Then, we can say "our car has linear motion". Also, we can see our car's velocity is changing ...

1

The car probably experiences a constant acceleration of $10{m/s^2}$. You can see from the chart that the velocity follows this as after every second the car is going $10{m/s^2}$ faster. However, this is clearly not the whole picture. We do not know the acceleration at 1.5 seconds, or 1.55 or 3.14. We can get confirmation that our acceleration model works ...

1

First of all don't insert actual numbers until the end. It makes it much easier to keep track and check whether your units check out. This problem is easier if you invoke conservation of energy. Simply equate: At t = 0. -Potential gravitational energy. At the end. -Kinetic energy of the toolbox. -Dissipated energy due to friction. You will find that ...

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