New answers tagged

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In your solution consider what happens when $t=0$. The projectile is at position $(0,0)$ and the jet is at position $(0,250)$ ie at a height of 250 m directly above the projectile. If this was the initially condition then the projectile would never hit the jet. As soon as you see a $\sqrt 2$ in an answer to this sort of problem you need to think about ...


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Projectile's path is a parabola, whose apex (point of maximum height) depends on $\theta$. For particular values of $\theta$, apex does not even reach up to the altitude at which enemy-jet is flying. So for all those cases where apex is equal to or greater than the altitude of enemy-jet, there is a $\textit{possibility}$ of hitting the enemy-jet (the gunner ...


-1

From an outsiders perspective, I see inertia as the state of the matter, and momentum as measurement of that state. Inertia can be a state of non movement or extreme speed. The rate at which this movement or non movement is allowed to continue is momentum. That is why momentum can be measured but inertia can only be observed.


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I think some errors crept into your calculation. And at any rate - your confusion is around $a_0$ - which is the quantity you need to solve for (the initial acceleration needed to reach the ball at just the right moment). The player has to cover a distance $d$ in the time that the ball covers a distance $\sqrt{2} d$. The velocity of the ball is $v_b$, so ...


0

Yes you could rise up the rope. You need to exert a force more than your weight, f = your mass x g. But you have to rappel exceedingly faster. Basically you could look at this as a rocket which is constantly refueling. If you exert less than your wait you break your fall to the degree of your force. You'd need lots of rope though depending on its mass.


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Take it one step at a time Time to reach $v_1$ under acceleration $a_1$ $$t_1 = \frac{v_1}{a_1}$$ Distance traveled to reach $v_1$ $$x_1 = \frac{1}{2} a_1 t_1^2$$ Total distance traveled during cruise (time $t_1$ to $t_2$) $$ x_2 = x_1 + v_1 (t_2-t_1) $$ Total time to stop from $v_1$ under deceleration $a_3$ $$ t_3 = t_2 + \frac{v_1}{a_3}$$ Total distance ...


1

For the horizontal portion, there is no acceleration, initial velocity is 12 m/s and final, I have no idea. See Newton's first law an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Your rock moving at 12 m/s horizontally has momentum in that direction. That momentum is ...


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Here I'll neglect air resistance (as well as assume there is no horizontal movement, as you say). Additionally, I'll say 'velocity' rather than 'speed', but since we're only moving in one dimension the velocity differs from the speed only by a sign. The initial velocity is provided by the act of jumping, or more precisely of pushing off the ground beneath, ...


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Knowing the final velocity $v_f = 0$ $m/s$, the acceleration is $a=-9.8$ $m/s^2$, and that the displacement is $\Delta x = 0.5$ $m$, you can use the kinematic equation $v_f^2 = v_i^2 + 2a\Delta x$ to find the initial velocity to be approximately $3.13$ $m/s$. Although the person starts at rest, the initial velocity is not zero, since they have to jump - and ...


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I agree with @Matt S's comment that you didn't state the problem because we can't see what is supposed to be happening in your diagram. Anyway, for every reasonable assumption I can make about what happens, the time of ascent is different to the time of descent. I assume your diagram shows a projectile in a friction-less parabolic trajectory that ends in an ...


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Here is how to answer this question (rather than the specific answer, which I don't think you're after). Firstly they tell you that the two bits of the path are the same length ('the first half of the path') and are straight. Call the length $l$. They also give you the angles, $\alpha_1$ and $\alpha_2$ of the two bits of the path to some reference line. ...


1

Consider what you stated: $$a=v\frac{dv}{dx}$$ Now rewrite it: $$\frac{dv}{dx}=\frac{a}{v}$$ If $v$ is tiny, then you know that $\frac{dv}{dx}$ must be enormous in order to produce the acceleration that you darn well know exists. Accleration at the top of that trajectory is surely $9.8$ $m/s^2$. As $v$ gets smaller and smaller as it gets to the apex, $\...


-2

The position vector of the particle is $\vec r(t)=x(t)\;\vec i+y(t)\;\vec j+z(t)\;\vec k\;.$ According to the definition, velocity and acceleration are as below: $$\vec v(t)=\frac{\mathrm d}{\mathrm dt}\vec r(t)$$ $$\vec a(t)=\frac{\mathrm d}{\mathrm dt}\vec v(t)$$ So, you just need to differentiate.


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Yes, it is possible to rise theoretically w.r.t a ground frame. But the rope-man system's center of mass must keep moving downwards because of the only external force acting on the system (gravity). The lighter $M$ gets, the harder it will be for the man to rise, and it will become impossible in the limit the rope becomes mass-less. This is intuitive because ...


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The question tells you that the speed is decreasing. It is NOT constant. For the speed to decrease, there has to be a force, and hence an acceleration, acting opposite to the initial direction of motion. This decrease in speed required a -ve acceleration. The question also says that the train turns round a circular bend. The centripetal force required to ...


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Problem statement says "the train slows down at constant rate", not "at constant velocity" - in fact, if velocity were constant the train wouldn't be slowing. What the statement means is that the train is slowing with a constant acceleration - at least, the tangential component of acceleration is constant. Since velocity is decreasing, the sign of the ...


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The problem doesn't say the train "is slowing down at a constant velocity." That doesn't even make sense --- either it's slowing down, and the velocity is not constant, or it's not slowing down, and the velocity is constant. It can't be both accelerating and at a constant velocity, since the very definition of acceleration is change in velocity with respect ...


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Because you can separate the vector space (assumption) of R x R into two separate vector spaces, each spanning the real line R with independent basis. When you are in the vector space R then you can't reach all elements in R x R. You can do that if you make a superposition of each of these R spaces into R x R and then you can reach all elements in this space ...


13

Your error is simply that you are assuming that $v(x)$ is differentiable with respect to $x$ at $v=0$. The chain rule needs that all derivatives involved exist before you can apply it. In the case of just letting go of something, the function $v(x) = \sqrt{2gx}$ is not differentiable at $x=0$, which is where $v=0$, so you are not allowed to apply the chain ...


3

When an object starts at rest, the change in velocity when it has made an infinitesimal displacement is infinite - in other words, $\frac{dv}{dx}$ is undefined. You can see this most easily by plotting the curve of $v$ as a function of $x$ for an object starting at rest: $$x = \frac12 a t^2\\ v = at\\ x = \frac12 a \left(\frac{v}{a}\right)^2\\ v = \sqrt{...


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This equation is best understood in integral form $$ \left. {\rm d}x = \frac{v}{a}\,{\rm d}v \right\} x_2-x_1 = \int \limits_{v_1}^{v_2} \frac{v}{a}\,{\rm d}v $$ It gives you the distance traveled by a varying acceleration between two speeds. "A car accelerating from 0 to 60 mph needs X distance". By stating that $v=0$ always not only it implies that $a=0$...


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The acceleration along the axis in this case depend upon the choice of axis. If you have chosen vertical as y axis then there will be only acceleration along y axis as it cannot have any component along x axis. I think you have problems with why it only affect the motion along the vertical. It is because tho orthogonal vectors doesn't affect each other as ...


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The kinematic equations are derived from differential equations. This means that you can start with the acceleration, then move to velocity, and then position given that the necessary information is available (like initial velocity or position). I find that understanding this helps me visualize the connection between the three. You can not rely on the ...


1

Why your computed average acceleration is wrong? the average acceleration is defined as: $\overline a=(v_2-v_1)/(t_2-t_1)$ where the $v$'s are instantaneous speeds. If you start with zero initial speed you can simplify it to $\overline a= v /t $ $v$ is still the instantaneous speed at $t$. For a constant acceleration $v=at$ so in this case you ...


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I want a more mathematical way to see it. Acceleration is defined as the rate of change of velocity with time. Velocity, being vector, can change just by changing direction keeping the magnitude constant. In a circular motion if the angular frequency $\omega$ is constant, then the magnitude of the velocity i.e., speed is constant but the velocity changes ...


1

If velocity at bottom of slope is all you want then energy method in Gert's and Dr Xorille's answers is an elegant and easy way to get at it. However you seem to be tangled up with resolving vectors. In working with vectors you should set up a coordinate system and stick to it until the end. If you have to set up more than one, and worse, have to switch ...


4

I find no trouble in thinking of velocity and acceleration vectors as arrows. First some definitions: Velocity is a vector. Speed is it's magnitude. Acceleration is a vector. It's magnitude has no new name. We agree that acceleration is present if there is a change in velocity: $$\vec a= d \vec v /dt$$ That is, any change. So if magnitude (speed) and/...


1

The glitch in your logic is that you supposed acceleration to be distance/time squared while using your formula: $$ \Delta x = v_0t+\frac{1}{2}at^2 $$ The above formula gives $$ a= \frac{2\Delta x}{t^2}$$ supposing that vo is equal to zero. This makes the acceleration equal to the average acceleration in your case since the car is moving in one direction ...


0

The equations you give are examples of the SUVAT equations, which are generally written as: $$\begin{align} s &= ut + \tfrac{1}{2}at^2 \\ v^2 &= u^2 + 2as \end{align}$$ where using your notation $s=h-h_0$ and $u=v_0$. Let's take your first equation: $$ h = h_0 + ut + \tfrac{1}{2}at^2 \tag{1} $$ The problem is that this relates the distance $h$ ...


0

No. Acceleration and velocity have different units, so their magnitudes cannot be compared. Whether one is larger than the other numerically depends on what units are used. Acceleration and velocity can have the same direction, but this is not necessary - eg a ball thrown upwards has upward velocity but downward acceleration, until it reaches maximum ...


1

Gravity acceleration is... acceleration, measured in $\mathrm{m/s^2}$. It is the rate of change of the velocity.. Velocity is measured in $\rm m/s$. It is the rate of change of the position. The vertical velocity and the acceleration due to gravity of a body are collinear but they can have different magnitudes as well different orientations. Think about a ...


-1

I believe I understand what you are asking and the answer is that the observer does not see an accurate representation of what happens near the event horizon. In the first graph the line would curve up to infinity at the end because as the object fell into the hole the light would be slowed to a stop at the event horizon. As the object begins it's approach ...


2

Let's consider the frame in which initially both the masses are at rest to be the frame $O$. In frame $O,$ momentum conservation is trivially followed because of the symmetry of the problem. For the energy conservation, we require that $M = m \sqrt{1-v^2}$, where $m$ is the initial rest mass of each of the particles and $M$ is the final rest mass of each of ...


1

You already have a relation between $\theta$ and $\alpha$. If this is correct, all you need to do is to rearrange it. If it is not possible to rearrange it to get the given relation, then either your equation is wrong, or the given equation is wrong, or both. If the particle grazes the upper plane at maximum height, where the velocity vector is horizontal,...


0

The other answers provide a good explanation of the situation. But,your observation might have been different if the car was constantly accelerating but the fly was flying at a constant speed.Then,it would have crashed with the back glass after sometime,as the car is moving forward faster than the fly is moving forward.



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