40
Suppose you are travelling at a uniform velocity and you cover 1 meter in 1 second. Your average velocity is
$$\frac{1\ {\rm m}}{1\ {\rm s}} = 1 \frac{\rm m}{\rm s}.$$
If you consider a 1 millisecond interval within that 1 second, you cover 1 millimeter. Your average velocity in that milisecond is
$$\frac{1\ {\rm mm}}{1\ {\rm ms}} = 1 \frac{\rm m}{\rm s}...
39
$$v_\text{average}=\frac{\Delta s}{\Delta t}$$
$$v_\text{instantaneous}=\lim_{\Delta t\to0}\frac{\Delta s}{\Delta t}$$
If the time interval gets infinitesimally small $\Delta t\to 0$, then you are dividing with something very, very tiny - so the number should become very big: $$\frac{\cdots}{\Delta t}\to \infty \quad\text{ when } \quad\Delta t\to0$$
If the ...
17
Given velocity $v(t)$, the distance moved after a certain time $t$ is not $v(t)t$ - this formula works at constant velocity, but when the velocity is changing, the correct expression is $\int^{t_f}_{t_0} v(t) dt$. Therefore your friend's third line is incorrect.
7
The error is just that $v(t)t$ is not the anti-derivative of $at$. This is easily checked by just taking the derivative.
$$\frac{\text d}{\text dt}\left(v(t)\cdot t\right)=v(t)\cdot\frac{\text d}{\text dt}(t)+t\cdot\frac{\text d}{\text dt}(v(t))=v(t)+at\neq at$$
It's a simple calculus mistake.
7
You're using $v$ in two different ways. In the first few equations, $v$ means the final velocity after a period of uniform acceleration, assuming one starts from rest. In the equation $v = s/t$, $v$ means the average velocity instead. That's half as much, which is why you're off by $2$.
This is a general warning for learning physics. In other high school ...
5
To give a purely qualitative answer consider the meaning of your friends third line
$$
x(t) = x_0 + v(t) \cdot t \;, \tag{1}
$$
(where I've made the multiplication explicit).
This claims that you find the position at moment $t$ by taking the initial position ($x_0$) and adding to that the elapsed time times the velocity the particle has at moment $t$.
So, ...
4
On the motorway, your passenger asks: "How fast are you going?". You look at the speedometer and answer: "109 km/hr". Had he asked 1 minute later, then you would have said: "111 km/hr". Had he asked 1 minute earlier, then you would have said: "110 km/hr". You are telling him the instantaneous speed.
But what if he asks you afterwards: "How fast did we ...
4
The displacement is only the velocity multiplied by the elapsed time if velocity is constant as you suggested. To derive the equation for varying velocity you must consider the infinitesimal case where the elapsed time is so small that you can consider velocity constant. In this case, a small displacement $dx$ is given by the product of velocity v(t) by ...
4
Why do we use velocity instead of speed for different physics
problems? I recognize how they are different but why use one over the
other?
My cottage is 100 km due north of my house in Toronto.
If I drive 100 km/h, will I arrive at my cottage in an hour?
What if I drive east?
What if I start in Montreal?
Thus velocity.
2
Your source should've worded that more carefully. This is not the acceleration of the particle in the rest frame; instead, it's the acceleration in a momentarily comoving inertial frame, that is, an inertial frame in which the particle is momentarily at rest. In the actual rest frame, which is a non-inertial frame, the velocity and acceleration of the ...
2
Let's say you are observing from your point of view two objects, traveling in their own directions and at their own speeds. So you have two velocity vectors $\vec{v}_1$ and $\vec{v_2}$.
By subtracting one from the other, you get the relative velocity between those objects: $\vec{v}_2 - \vec{v}_1$ will be the velocity of object 2 as observed by object 1 (...
2
I took another stab at it and I was able to solve it using the following equations: $$x_f=x_i+v_xt$$
$$y_f=y_i+v_y{_i}t +\frac 12at^2$$
Applying and rearranging these to my problem
For x-direction I get:
$$d=0+v_i\cos\theta\cdot t$$
$$v_i=\frac{d}{\cos\theta\cdot t}$$
And for y-direction:
$$0=h+v_i\sin\theta\cdot t -\frac 12gt^2$$
Apply previous equation and ...
2
Consider a streamline along which the fluid velocity is increasing. This means that the fluid is accelerating, so there must be a force. Forces in an ideal fluid are related to differences in pressure, so the pressure must be decreasing.
2
You had the correct tangential, $c$, and radial, $\dfrac {c^2t^2}{b}$, accelerations
If the radius is constant then in polar coordinates $\vec v = b\,\dot \theta \hat \theta = c\,t\, \hat \theta \Rightarrow \dot \theta = \dfrac{ct}{b} \Rightarrow \ddot \theta = \dfrac cb$ and $\vec a = -b\,\dot \theta^2\, \hat r + b\,\ddot\theta \,\hat \theta =-\dfrac{c^...
2
I think I understand it now, mathematically speaking, but is there a
more conceptual answer?
OP evidently seeks a conceptual answer to why $x(t) \ne x_0 + v(t)\cdot t$ when $v(t) = v_0 + at$ and $a$ is a constant.
Consider the simple case that the initial position and initial velocity are zero. Stipulate that $v(t) = at$ where $a$ is a constant and it ...
2
You're confusing velocity with acceleration. The acceleration "a" acts on the stone only until you let go.
"After the release" means "after you let go", hence you no longer impart any acceleration to the stone, and "a", the acceleration of the lift, no longer affects the stone. In your balloon experiment, the stone has velocity "u", but in the lift the ...
2
I think you are querying the necessity of the statement
Infinite accelerations are not allowed
To get zero displacement one has to have a positive displacement and a negative displacement and hence at some stage the velocity must be zero as shown on the left hand graph which is continuous and well behaved.
Now what about going from a velocity $...
1
In part. This is the equation of position as a function of time for uniformly accelerated motion with $t$ shifted by $t_0=2s$, i.e. in terms of $(t-t_0)$ rather than just $t$.
Note that the velocity at $t=t_0=2s$ is $u$ but that’s not the initial velocity. Since the acceleration is constant with $t$, you can read it off at any $t$, including a $t=t_0=2s$, ...
1
So the answer should be given by taking a particular frame of reference....imagine if you are in the elevator, you will feel that the acceleration of the stone is g+a downwards but now imagine that you're on the ground looking at the stone falling, you will feel that the acceleration is simply g downwards hence I think that the answer was given taking ground ...
1
Your teacher is right. You are confused about acceleration and velocity, which can be tricky.
Acceleration is the change of velocity with respect to time and nothing can be accelerated without a force acting on it. Think about it. Your car won't drive unless the engine is powering it. Your bike won't go unless you pedal. So as soon as a force starts/...
1
You’re right. You’ll get the change in position $\Delta x$, which is the displacement. Maybe your teacher is confused about “displacement” ($\Delta x$) vs. “position” ($x$).
1
Each derivation rests on the assumptions used. The standard kinematics equations you mention first, depend on the assumption of constant acceleration.
My problem is that your friend hasn't stated what assumptions he used to get to $x(t) = x_0 + v(t)\,t$.
So let us differentiate both sides to see what kind of acceleration is needed (using the product rule)
...
1
You have a few misconceptions. The friction which you draw in the AB direction actually acts in the opposite direction, toward the center of the curve. This friction component is the force component which has magnitude of precisely $mv^2/r$. That $mv^2/r$ value simply tells the magnitude of force necessary radially in order to travel a certain curved path at ...
1
Acceleration is the derivative of velocity with respect to time, by definition.
It doesn't matter what factors affect it, that's still what it is. You could have acceleration as a function of the amount a spring is stretched, acceleration as a function of how much you press the gas pedal, acceleration of a sail boat as a function of how fast the wind is ...
1
Circular motion at constant speed had varying velocity because of the constantly changing direction of motion. This made possible the evaluation of centripetal force causing the rotation which depended on the rate of change of velocity( direction was changing) not speed. Had it not been so we would not have been able to determine the force(cause) behind ...
1
The trajectories of the thrown object and its consecutive bounces will be parabolic in their shape. In this case the ball will lose some of its kinetic energy each time it make contact with the Ground. However, the angle it makes with the ground each time will be equal to the angle it was original throw at. This means that all of the trajectories will be ...
1
Will this car slow down?
In the ideal where the tires have sufficient grip, no it doesn't need to slow down. In reality, the tires must slip a bit while turning, and this slip will introduce some energy loss that is not present in the straight-line driving. But that could be minimal.
What is the actual physical reason that causes cars to slow down in ...
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