# Tag Info

## New answers tagged velocity

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As it turns out, adding mass (while keeping the dimensions of the car fixed) will make the car go faster. This sounds contrary to what all physics students are taught, but the reason is while friction scales with mass, air resistance doesn't. That's why a thirty-foot rock will fall faster than a thirty-foot parachute in an atmosphere. It's also why the ...

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Try applying conservation of energy, $E_i = E_f$, as the problem suggests. To start, you know that the ball initially has kinetic energy and gravitational potential energy but ends with only kinetic energy.

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The force due to sliding friction causes the center of mass to decelerate - since it is the only external force acting on the ring, and we know the center of mass moves as though all forces act there. You might find this earlier answer of mine and the links therein helpful. As for the "in the real world the ring will stop completely" part of your question - ...

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I've no idea what a "force match" is, but the numbers are pretty clear. You seem to have missed the fact that, since there are 4 wheels, the total decelerating force will be 4 times the force per wheel. In this case, that means that the total decelerating force is 2800 N. a = F/m = 2.8 m/sec^2 Furthermore, 50 km/hr is 13.888 m/sec. so t = v/a = ...

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The velocity addition formula you cite does not quantify what you think it quantifies. I'm going to say see and when I do, I mean "computes relative to its own frame". The velocity addition formula describes the following setup. Frame 0 sees a particle (particle 1) moving in a direction at speed $v_1$. Frame 1 sees that particle 1 at rest and frame 1 ...

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You would have a problem withe a good definition of the start. If I had such a need (to measure my hand velocity 10ms after a start) I would prepare two wires with lengths difference about 10cm one end attached to my hand, the other two ends attached to two contacts of some device. 1/ arduino or something similar - you could measure a difference between the ...

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Attaching an accelerometer onto your hand and linking it up to a computer that will record it's live data could work. Sadly, I have little idea about what sort of software might be good to analyse the data. I get the feeling that arduino might be useful for such an application.

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Let's say you have 3 systems. $B$ moving relative to $C$ with velocity $u$ and $A$ moving relative to $C$ with velocity $v$, all along one axis. $A$ will "measure" for the velocity of $B$: $$u' = \frac{u-v}{1-\frac{uv}{c^2}}$$ While $B$ will "measure" for the velocity of $A$: $$v' = \frac{v-u}{1-\frac{uv}{c^2}} = -u'$$ It holds true that the velocity ...

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You must solve for the objects initial velocity first: $$v(t)=u+at\\ v(0)=u\\ v(1)=u+10\text{m/s}\\ =20\text{m/s}\\ u=10\text{m/s}$$ With this adjustment you should find the correct answer.

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If the object was released from rest ($u=0\,{\rm m/s}$), what is its speed after $1\,{\rm s}$ if $a=10\,{\rm m/s}^2$? Using: $$v=u+at$$ you will find that the object was not released from rest...

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A nice way to compare both is to invoke the definitions: $${\vec a}_{\rm avg} = \frac{\Delta {\vec v}}{\Delta t}$$ and $${\vec a}_{\rm inst} = \lim_{\Delta t \rightarrow 0}\frac{\Delta {\vec v}}{\Delta t} = \frac{d{\vec{v}}}{dt}$$ Graphically, and if you consider change over an infinitesimal time period $\Delta t \rightarrow 0$, the same definition ...

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The instantaneous acceleration is the time derivative of the velocity vector: $$\vec{a} = \frac{d\vec{v}}{dt}$$ If the velocity is changing then the acceleration will be non-zero.

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Speed has no direction while velocity does. For example, if I say that I'm running at 10 mph, I have given you my speed. If I say that I'm running 10 mph north, then I have given you my velocity. Acceleration is the rate of change in velocity. Imagine this: I am in my car and you look at me before I even press the gas pedal. You close your eyes then open ...

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The effects of inertial acceleration is best understood from the physics of Newtonian mechanics. A good site to support your understanding of this physics: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html I recommend studying the trajectory calculator as it applies to your problem.

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Yes you can! For example: when you are sitted, doing nothing. From the reference system of the earth you are firm, with no velocity BUT there's the gravity acceleration. We can say that we have everytime a potential to have a velocity if we have a force. For the reason that says that for every force there's an other force with same value but with a contrary ...

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The problem is that 9/13 is approximately 0.692. You used decimals, the book used fractions.

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