# Tag Info

1

Use the standard equation: $$s=\frac{v^2-u^2}{2a}$$ break the problem into two halves: one accelerating and one decelerating. Taking the sum of both equations you should get the total distance traveled as a sum of some terms, all of which you know except the speed in the middle which should cancel out (assuming magnitude of $a$ is equal in both cases). ...

1

If you impact the second body its axis of percussion it will purely rotate. By carefully choosing the inertial properties of the two objects you can make the first object stop translating in the process. See this post for more details on a particle to rod impact.

0

Like New_new_newbie said, the question asks you to find the distance travelled in its third second (from t=2 to t=3 seconds). So just find x(2) and x(3), and the difference should give you 29m.

0

See this site: http://www.physicsclassroom.com/class/1DKin/Lesson-1/Speed-and-Velocity Example:In everyday use and in kinematics, the speed of an object is the magnitude of its velocity (the rate of change of its position); it is thus a scalar quantity.The average speed of an object in an interval of time is the distance travelled by the object divided by ...

3

There are many reasons why acceleration would not be constant. Books often don't mention it because they are getting mathematicians used to the concept of first derivatives before moving onto second, third etc... Consider the following example which ends up leading to Tsiolkocsky's Rocket Equation: http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation A ...

2

This does exist, and it is called Jerk, see the Wikipedia page on jerk. It is used quite frequently in physics concerning humans, as we are able to sense this, and there are limits to how much jerk a human can endure. It is, however, quite abstract and therefore more difficult to comprehend, which might be the reason that lower level textbooks do not ...

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Your question is not weird, is legitimate. It is possible, it exists, can be of use and it is called jerk, jolt, surge or lurch, and is defined by any of the following equivalent expressions: $$\vec j(t)=\frac {\mathrm{d} \vec a(t)} {\mathrm{d}t}=\dot {\vec a}(t)=\frac {\mathrm{d}^2 \vec v(t)} {\mathrm{d}t^2}=\ddot{\vec v}(t)=\frac {\mathrm{d}^3 \vec r(t)} ... 0 If you want to go from a to v you have to integrate over time:$$a=\dot{v}=\frac{d v}{dt}\int a dt =\int \dot{v}dt=\int \frac{d v}{d t}dt=\int dv = v$$0 You are almost done. If you simplify the express you found for v_{avg} and compare it with the initial equation for v_{avg}, in which you have to substitute v_i for v(t_i) and v_f for v(t_f), you will see that they are equivalent. Another way of deriving this is by expressing v(t) in terms of v_i and v_f:$$ v(t) = v_i + \frac{v_f - ...

1

There is one error in the derivation, if you want to have $v(t_i)=v_0$, you must have $$v(t) = v_0 + a(t-t_i)$$ You also have to use the fact that $v_f = v(t_f)$. Once you use all this, you should be able to divide out $t_f-t_i$ in the corrected version of your last line and get the result you seek.

4

You are free to pick your frame of reference. You can point $y$ up or down, or even sideways. You can put the origin at the top of the roof, at ground level, or at the center of the earth. My recommendation - in problems like you are describing, ALWAYS draw a diagram that shows what conventions you use - after that, you essentially answer your own question. ...

1

It depends on what my physics teacher might call your "direction of reference". Sure, you can make "down" positive. In this case, there shouldn't be any preferred direction along the $y$ axis - down is just a valid a positive direction as up - that is, as long as you adjust the other variables to match. For example, if the acceleration is $g$, you would say ...

0

You are asking for Galilean formula for adding velocities(since boats speeds are not relativistic ;) ) You always add the speeds but in vector form. This means that you will get a minus if they are opposed and a plus if they are in the same direction. But if they act on different lines, you need to find the third side of the triangle formed having the two ...

0

Hint 1: Everything accelerates (the velocity increases) downwards by 9.81 m/s, every second - on earth. Hint 2: The relationship between height and velocity can be calculated using kinetic and potential energy equations: 0.5*m*v^2 = m*g*h m = mass, v = velocity on the ground, g = gravity (9.81) and h = initial height. Since you know the velocity ...

2

Because that is where the lines normal to the wheels and through their centers intersect. If the center of rotation was at any other point, then there would be a component of velocity perpendicular to each wheel (wheel slip) and ideal wheels roll without slip.

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KERS systems (as used in F1 and leMans cars) rely on a flywheel to store kinetic energy. When braking, the transmition is clutched trough a 1:n gearbox to the flywheel. That is, one full revolution of the car's wheels will make the flywheel turn n revolutions. The flywheel system will then store the energy until it is needed. Since a flywheel system rotates ...

1

If I understand your question correctly you are asking why you should use $t_2-t_1$ in the denominator instead of $t_2+t_1$. The reason is that both measurements are taken with respect to some initial time. This initial time is arbitrary; think about it as the time when you started your stopwatch. The total time elapsed between the events occurring at ...

2

I just figured out the answer on my own so I'll post it here for the sake of the site. $$\Delta x = v_0 t \cos \theta$$ $$\Delta y = v_0 t \sin \theta - \frac{1}{2}g t^2$$ Where $\theta, \Delta x, \Delta y$ are all known. Using the first equation, we get $v_0t = \frac{\Delta x}{\cos \theta}$ Now substituting into the second, we find $\Delta y = \Delta ... 2 There exists no case where radial acceleration does not change direction. But i thought of a quite eccentric stuff like this where it appears that it does not change direction.Suppose as the ball tied to the string revolves,there is an observer who is moving along with the string facing the ball.Then, in that frame of reference the direction does not ... 3 This is the diagram you need to draw: The red arrow is the net velocity of the boat - the sum of 25 km/h going due North, and a current of 10 m/s at 60 degrees East of South. (not drawn to scale) Just do simple math to determine the size of the two green segments - then compute the angle (heading) from the arctan of their ratio. 2 You can't calculate the displacement like this. The application note clearly says " When implementing positioning in 3 axes, extra processing is required to null the earth's gravity effect.", and that's the understatement of the month! Nulling gravity is an enormous problem, unless your accelerometer is perfectly perpendicular to Earth's gravity vector! The ... 1 In this case distance and displacement mean the same thing. Just calculate the area under the second graph as you did for the first one. The difference between distance and displacement arises when the direction you're moving in changes. For example suppose I travel one mile north then one mile south - obviously this brings me back to where I started. The ... 0 Although it's a little bit hard to see, the second graph axes are time (tiempo) on X, and velocity (Velocidad) on Y. You should be able to compute the position of the object after 5 seconds (it's not clear whether the velocity is supposed to be$0$at$t=0$, or$5 m/s$- the axis is not labeled at that point - but I expect that you have been taught how to ... 1 You have a few basic formulas for solving this kind of stuff: $$\vec{x}_f=\vec{x}_i+\vec{v}_it+\frac{1}{2}\vec{a}t^2,$$ $$\vec{v}_f=\vec{v}_i+at.$$ These are vector formulas, but all you're doing with the vectors is adding/subtracting them. When you add vectors you add the individual components, i.e. $$\vec{a}=(a_1,a_2,a_3)$$ $$\vec{b}=(b_1,b_2,b_3)$$ ... 0 You want to define the equation of motion by applying constrains to a model, in your case a second order polynomial. A second order polynomial has three degrees of freedom, namely$x_0$,$v_0$and$a$for the following function of time, $$x(t) = x_0 + v_0 t + \frac{1}{2}at^2.$$ I do assume that initial velocity implies the velocity at$t=0s$. The ... 1 It is a matter of sig figs. I had the same problem (while teaching a class). If you carry out the numbers far enough, you will get consistent answers. I assume one of the variables you had to originally solve for which is why after you use it to solve for another, you get different answers. If your solved variable is not precise enough, the error will be ... 0 First, you have to find the maximum height of the ball. In order to do so, we use the condition that$v_{y}=0$. $$0=v_{0}\sin(\alpha)-gt_{H}$$ From this, the time to reach the maximum height$t_{H}$is $$t_{H}=\frac{v_{0}\sin(\alpha)}{g}$$ From the vertical displacement, we obtain the maximum height using$t_{H}$... 0 In projectile motion the horizontal velocity is always same through the journey, only vertical component of velocity changes. After resolving the given velocity vector U (say) into X and Y components as Ux and Uy respectively. You can write R=Ux.T ( R is horizontal range, T is time of flight) Therefore, R=u^2 sin(2 theta)/g…(1) Here theta is the angle ... 5 When observing supernovae, often the bulk of the material is optically thick, so we only see the surface layers. These are moving at some speed$v_\mathrm{surf}$. When modeling supernovae, the very simplistic model everyone loves to use is that of homologous expansion of a uniformly dense (density$\rho$) sphere. That is, velocity of the material increases ... 4 No, in general, viscosity and density are not related. For example, for gases, we have: Maxwell's calculations show that the viscosity coefficient is proportional to the density, the mean free path, and the mean velocity of the atoms. On the other hand, the mean free path is inversely proportional to the density. So an increase in density due to an ... 0 KERS is also known as regenerative braking. It is a system found on the latest hybrid cars. When a vehicle is stopping, the brakes are applied to the discs to generate friction to slow it down. This generates heat which is energy that is lost or wasted. Regenerative braking creates friction in addition to the brakes, however, some of the energy is ... 1 Displacement is $$x = \frac12 a t^2$$ Without solving for acceleration, the ratio of displacements for two times is $$\frac{x_1}{x_2} = \frac{t_1^2}{t_2^2}$$ So if$x_1 = 5$, then it follows from the above that$x_2=20$and the difference (the distance covered in the second time interval) is$15 m$. 0 If it undergoes constant acceleration$a$from a stop, then at time$t$seconds it will have traveled${a t^2}/2$distance. So substitute: if${a 1^2}/2$equals$5$meters, then$a$equals$2 * 5$equals$10m/s^2$. You take it from there. 1 In the question, you clearly stated it takes 4 seconds for you to hear the rock hit the bottom of the well In 4 seconds, a rock (ignoring air drag) would drop$\frac12 9.81 (4)^2=78.5 m$- but it would take sound about 1/4 of a second to reach your ear, so the well must be a little bit less deep (1/4 second is almost 10 meters at the velocity near the ... 0 Last years I took regular physics and learned that distance equals Velocity x Time. That's not quite correct. In one dimension, the distance travelled equals the average velocity multiplied by the change in time $$\Delta x = \bar v \cdot \Delta t$$ where$\bar v$is the average velocity. If the velocity is constant, we can write $$\Delta x = v ... 2 One of the five most common equations of motion in high school physics is$$\Delta d=v_0\Delta t+^1\!\!/_2\,a\Delta t^2$$v_0 is your initial velocity (0 for your case), a is the acceleration (g for your case) and \Delta t is the time. That's the easiest equation you'll find for something like this, from here they get a bit messier. 3 Because of the homework policy, i'll just remind you of the definitions.$$Average \hspace{1mm}Speed= \frac{Distance \hspace{1mm}Traveled}{Time\hspace{1mm} of\hspace{1mm} Travel}Average \hspace{1mm}Velocity= \frac{Displacement}{Time}$$With this, you should be able to answer your own question. 5 This is unphysical. There is no force or momentum that would make something unstoppable. First of all, any moving object can be made to diverge from a straight line path simply by applying a small force perpendicular to its motion. Secondly, no matter how large a momentum an object has, it can be completely halted by a similar object with the same momentum ... 2 Well let's try and derive it:$$v=u+\langle a\rangle t $$holds because your professor says so. Now try and integrate:$$ s=ut+\int\langle a \rangle t dt $$If a is a constant that we get s=ut+0.5at^2 and combining with v=u+at to eliminate t we should get v^2=u^2+2as. But, if a is not a constant then \langle a\rangle is a function of time ... 2 Whenever you're confused about how to calculate some quantity, try going back to the definition. Think carefully about what the definition of average acceleration is:$$\vec a_\textrm{avg}\equiv\frac{\vec v_\textrm{final}-\vec v_\textrm{initial}}{\Delta t_\textrm{elapsed}}.$$Which velocities does this equation depend on? Which velocities does it not ... 2 Your result is correct: 1.351808043257857384251436296045? and you can get the length of the track in a very simple way$$ L_t > 1000m *\frac{148.15}{147.95} > 1001.351808.... m$\$ That is the bare minimum, they gave you a round figure to be on the safe side, that's all EDIT: The digits are mathematically accurate, but physically meaningless ...

2

Not sure I understand your question. Let me know if the answer doesn't make sense. When a accelerometer is in a given orientation to the direction of gravity, then the value measured on each axis is the dot product of the gravity vector and the orientation vector of the sensor. Imagine that you start with the sensor in the "normal" orientation: X points in ...

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While I would normally be the first to argue that people use too many significant figures, and leave it at that - I think there are two important points worth noting here: First - the times are given with a precision of 1/100th of a second over almost 150 seconds - we should take that into account when formulating our answer Second - when you say "how big ...

0

1.351 would round up to 1.4. You have enough decimal places in there to get your measurements into the subatomic range, +/- 10cm is accurate enough for a 1km olympic competition. Turn to the chapter on Significant Digits in your textbook. In short, don't include more decimal places than the least precise value in the entire process. To generally keep your ...

2

Why? Because you did not answer the question as stated... The question explicitly says that the time spent at each speed is the same. You specifically started by saying the distance at each speed is the same, and along the way you find that the time at each speed is different Different question... Different answer... The best way to find the average ...

2

Your approach, though original, does not yield the average. There are a few important factors here. The car travels each speed for the same amount of time Those speeds are constant (there is no speeding up/slowing down) Due to factor #1 being true, we don't have to worry about weighing the average. We can take this average in the simplest way: average ...

2

Probably there are many ways to approach the problem, The simplest, the first that comes to my mind is : find the time it takes. Think of the time it takes A to cover/catch up the difference in space using the difference of speed. P.S I hope this hint is not against the rule, if it is, just tell me and I delete my answer.

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The "missing" kinetic energy is still there ... it's now in the rotation of the yo-yo.

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This is kind of hard to describe without a diagram but anyway i will give it a shot. I am assuming you mean something like the fourth photo on this page: http://www.oldschool.com.sg/index.php/module/PublicAccess/action/Wrapper/sid/9595afb87c8cf767f034c3ae53e74bae/coll_id/4745/recs_ppg/5/desc/wrap-function/pg_id/3 If you consider the simpler case where the ...

1

The plane of the earth's orbit is extremely stable. Of course, the earth's orbit is affected by the other planets, especially Jupiter, but all the planets orbit in approximately the same plane, so the forces pulling the earth's orbit out of its plane are small. We can see that the planes of the planets' orbit are stable, because all the planets are in ...

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