# Tag Info

## New answers tagged kinematics

0

Can you have a zero velocity and nonzero average acceleration? If you by velocity mean instantaneous velocity, then the question makes no sence. The corresponding acceleration will as well be instantaneous (and the answer would be yes.) $$\lim_{\Delta t \to 0}a_{av}=a_{inst}$$ ... unless it is not a requirement that the acceleration is averaged over ...

2

For a particle in a gravitational field treated as a constant? Surely Newton's equations of motion in the fixed rectangular frame: $$\ddot{x}=0$$ $$\ddot{y}=-g$$ are as simple as it can get!

0

If the range is not too large, you can approximate the acceleration (due to gravity) vector to have the same direction over the trajectory of the projectile. It will probably be a good idea to use earth fixed 2D cartesian coordinates in this case, one axis horizontal and the other vertical to the Earth's surface. The acceleration acts only along the vertical ...

1

Initial kinetic energy is $K_1=\frac{1}{2} m (v_x^2+v_y^2)$ with potential energy $U_1=0$. At the apogee, the potential energy is $U_2=m g h$ and the kenetic energy is $K_2=\frac{1}{2}m v_x^2$. Equate the two sums to get your answer. $$U_1+P_1 = U_2 + P_2$$ $$0+\frac{1}{2} m (v_x^2 + v_y^2) = m g h + \frac{1}{2} m v_x^2$$ $$\frac{1}{2} m v_y^2 = m g ... -2 Where U is the intial velocity,g is the gravitational constant,R is the range of the projectile and ∅ is the launch angle between the projectile and the flat ground (imagine the flat ground as the x-axis with the projectile inclined at an angle ∅ on it). 0 If your direction of motion is linear it does not necessary mean that the equation representing its position with respect to time is always linear.And if the velocity is constant it always means that acceleration is zero. 4 Yes, because acceleration also includes change in direction. For example, a race car on the track goes in a circle. If its speed is 150 mph for the entire race, it is still accelerating because it is not going in a straight line. 2 1) On integrating dt on the RHS we get a +c(constant of integration) but why is there no +c on the LHS while integrating dv? If we start with:$$ dv = adt $$and integrate both sides then we can indeed have a constant of integration on both sides:$$ v + C_1 = at + C_2 $$but we can just subtract C_1 from both sides to get:$$ v = at + (C_2 - C_1) = ...

0

In the first step ( $\int dv=a\int dt$), we get $v+c_1=a(t+c_2)$. and therefore $v=at+(ac_2-c_1)$. The term $ac_2-c_1$ is constant (because $a$ ,here, is constant). Instead of tediously writing the integration constants for every step, we can write one at the ending, since the last constant ($c$, in this case) absorbs all the other constants which have ...

-1

It's an easy method. You need to calculate m/s, when you have distance in meter and time in seconds. Just use the simple relation. 1 minutes equals 60 seconds. That means 320941.03 minutes equals 320941.03*60 s. Which is 19256461.8 seconds. Now use the formula, speed equals distance per time. i.e, v = s/t = 7720298/19256461.8 = ...

0

To find the speed in m/s, you will need to have a distance measurement in meters (which you already have), and a time measurement in seconds (which you don't). You can change a measurement from one unit to another version of that unit by multiplying by 1. Multiplying a number by 1 doesn't change the value, which is why changing units doesn't change the ...

1

If you take the ratio of the values you have, you will have the velocity in meters per minute, right? Then, to obtain the value in meters per second, you just have to convert that time from minutes to seconds. You can do that by multiplying by 60, because 1 minute is equal to 60 seconds. Therefore, your time is now $320941.03 \text{ minutes} \times ... 0 Key to understanding this question is the fact that a "football", in this case, is a hand-egg: When it is given spin about its long axis, it will act a bit like a gyroscope: it tends to maintain its attitude. So the question becomes - when is that desire of the ball to keep pointing in the same direction greater than the torque due to airflow that ... 1 As the particle travels around the cylinder (having radius r) and the taught string wraps around the cylinder, the path of the particle will trace out a spiral with radius that decreases as a function of the cylinder radius. In other words, the radius of the spiral (let's call S) equals the string length at any given point. All you need to do is find the ... 0 The path will be an Archimedean spiral. I think the key to this is that the speed of the particle does not change. You have an equation for the angular velocity in terms of the radius. You also have an equation for the rate at which r is changing in terms of its angular velocity: r is changing by $$-2 \pi R\omega$$ Which seems to be all you need. 0 I'm shocked that there isn't a satisfactory answer on this site yet! This can be answered in any number of dimensions with some relatively simple vector math. As follows intuitively (rigorously in the center-of-mass frame), for equal mass particles, the relative velocities of the particles are reversed along the normal direction. All that needs to be done ... 0 I just need to check if my understanding of the transfer of gravitational potential energy to kinetic energy is correct. Is anything wrong below? Yes. Gravity is not a force in the Newtonian sense. The potential energy of the body isn't stored in the gravitational field or the Earth, it's in the body itself, in its mass-energy. When you drop the body the ... 0 The velocity you have correctly written is independent of the mass of the object, hence any object of same mass follows that velocity law. However, if you count in the drag force then you must add in the drag factor, meaning your velocity of the falling object now would be $$v=(\frac{2mg}{\rho AC})^{1/2}\tanh[t(\frac{g\rho A C}{2m})^{1/2}]$$ where$\rho$... 0 Yes, two objects with the same shape but different masses will have the same velocity after being dropped from the same height. Galileo tried this a while ago by dropping two such objects from a tower. Factoring in drag, we know that drag ∝ velocity2. So if the object (falling from the same height) has a greater surface area/size, its drag will be ... 0 Using gonec's suggestion, we can solve for$k$. Now, taking the sign convention that left is negative and right is positive, lets consider the problem. In simple harmonic motion, the position is determined by a sinusoidal function,$x = x_{max}cos(\omega t + \phi)$, taking the equilibrium position to be$x=0$. Now, assuming the particle to be connected to a ... 0 The parachute can not be opened instantaneously. The parachute is fully opened at the "vinicity" of the point$C$. The acceleration due to gravity is constant but there is another force one needs to take into account : the air resistance. 1 You are correct that centripetal force does not affect speed. So let us consider tangential speed, which is the speed the turning car would have if centripetal force were removed. Linear speed = tangential speed = distance / time = (2 * pi * radius) / time. (To simplify, assume the car goes through one full circle at uniform speed.) The linear speed of the ... 0 I would say that indeed the center of the car still move at 50km/h. The angular velocity is$\omega = v/r$, with v the tangantial velocity and r the radius. Therefore the wheels on the inside will go slower (smaller v) and the wheel on the outside will go faster (larger v) to keep the same angular velocity$\omega$. Best, Samuel 2 In reality you have$a(v) = g - \beta v$and you want$v(t)$and$x(t)$. This can be achived with direct integration, instead of coefficient matching and diff. equs. $$t = \int\limits_0^v \frac{1}{a(v)}\,\mathrm{d}v= \int\limits_0^v \frac{1}{g-\beta v}\,\mathrm{d}v = -\frac{1}{\beta} \ln \left(1- \frac{\beta v}{g} \right)$$ or $$v(t) = \frac{g}{\beta} ... 1 To answer your question directly, when the direction of the velocity of the front of the car does not match the front wheels' direction, the tire is deformed, acting like a spring which exerts a lateral force on the car. The deviation between these directions is called the slip angle, and to a first order approximation the lateral force on the tire is ... 1 Without any ramps that would convert forward momentum into vertical momentum or vice versa the forward movement can be ignored. The gravitational potential energy from the fall must therefore be absorbed during the collision as the final vertical kinetic energy is zero.$$G.P.E. = m\,g\,h$$Where m is that mass of the bike and rider, g is acceleration due ... 1 The definition of acceleration, a = \frac{dv}{dt} is always correct, and any other equations of acceleration are derived from it, depending on the conditions. To use this equation, you need to have a basic understanding of calculus. For example: condition: constant acceleration. This means \frac{dv}{dt} is constant. This means that v = at + constant ... 0 when you accelerate the vehicle and you turn, the center of gravity shifts from different parts of the car body. Maybe I am misinterpreting your question though.. 0 The equation from the link is an approximation. In your analysis you do not include the mass of the car when considering linear acceleration. Nor do you consider the reaction force of the wheel/axle accelerating the car. You can split the driving torque into two portions, the portion that rotationally accelerates the wheel, and the portion that is ballanced ... 2 Definition As in the question we'll define our bicycle as the location in two dimensions of our contact patches r_1 and r_2 for the rear wheel and front wheel respectively. All other aspects of the bicycle will be ignored. Assumptions The first assumption corresponds to the distance between the wheels being constant:$$|r_2-r_1|=C'$$... 0 Other answers and the referenced paper assume a constant radius turn. This path would require a discontinuous steering angle, which is not only non-physical but a somewhat poor approximation for how human drivers drive. The reason drivers don't approximate this technique (besides that it would take very rapid steering wheel movement) is that it would incite ... 1 No they won't. while they may be circles that are centered on the same point, the radius of the four circles is different. This diagram ought to convince you: The four wheels trace out circles with radius r_1 ... r_4, and each of these will be different (in general). It is possible, for certain combinations of L, W and D that the front right and ... 0 Consider a mass which is dropped from a height. This is done in a vacuum, or is otherwise shielded from crosswinds. Furthermore, the mass is located on a non-rotating body, has no lateral initial velocity, and is not noticeably affected by the gravitation of any other bodies. Two things are clear: first, the mass moves in a straight line toward the center ... 3 You are using the word "linear" in two different ways. When an object moves along a straight line we can say its motion is linear - but that does not mean its acceleration is zero. Just that the acceleration points along the same direction as the velocity (so no change in the direction of the motion). The second meaning of "linear" is in the exponents of ... 0 Velocity is a vector quantity having both magnitude and direction. Change in either one of these or both is acceleration. A body moving along a straight path would have constant direction but might have variable speed. 3 You should always apply energy conservation, and it ought to hold in all reference frames, including the frame in which the sigma is at rest. In the sigma's rest-frame,$$ E_{\text{initial}} = E_\Sigma = m_\Sigma $$and$$ E_{\text{final}} = E_\Lambda + E_\pi \ge m_\Lambda + m_\pi $$Thus we have that,$$ E_{\text{initial}} < E_{\text{final}} $$The ... 1 Energy conservation always applies. Your mistake is in thinking that adding masses will solve the problem instead of clarifying some aspects. In the case of the sigma-zero the decay at rest allows to see that the sum of the constituent masses is larger than the mass of the sigma-zero. For a decay to happen there should be energy left over to go to the ... -1 You should remember that photons don't have any mass, so relation is E=p/c for photons. Moreover, you have to count electrons mass when you apply energy equations. Electron mass will vanish, but it's important to take account of it. I mean, before the square root, you don't have the electron mass, you have the kinetic energy. 14 There are several interesting things going on when a car turns. First - let's take the simple diagram of two front wheels turned by 45°: As you can see, the top tire would like the car to turn around the point C_1, but the bottom tire (at the same angle) wants to turn around C_2. This means that in reality both tires will experience some lateral ... 2 Arrows and paintballs go about the same speed, 200-300 feet per second. In episode 109, Mythbusters confirmed a professional ninja can block an arrow with a sword. It's plausible. However, distance is going to make a difference. I don't have the episode to check how far away the archer was, but in your question 24 feet (8 meters) is really close. It ... 0 As always in this sort of subject, you should consider Mythbusters, not just mention it. https://www.youtube.com/watch?v=2qY0DA7DG9s And the answer is pretty much, no you can't. It's reaction time that's the problem. 2 I don't understand why this question has not been answered in the comments. If there's no friction, then the "direction of the wheels" cannot turn the car. It's as though the car were on a sheer-ice skidpad. (Those of us who have driven in winter weather know the feeling of encountering a patch of "black ice".) 7 Correct me if I'm wrong, I believe that from a paintball gun when shooting there is no smoke, vapour or other visible exhaust and not even recoil. No visible signal. The dodger must rely on sound. Sound reaching the ears. The speed of sound in air is around v=330\:\mathrm{m/s}. Sound will propagate the x=8\:\mathrm{m} and reach the dodger in ... 2 The time that the person has to dodge is t=\frac{8 m}{160 m/s}=0.05 s So he has to move at a minimum speed of v=\frac{0.50 m}{0.05 s}=10m/s 2 That is partially correct. You are missing a critical part of Newton's 2nd law: F_{net} = m a. It's not just any one force that equals the mass times the acceleration. It's the "net force" - the vector sum of all the forces. So a car that is traveling at a constant speed may have many forces on it (gravity, the normal force, friction, air resistance, ... 1 I am going to make some general comments, expecting that they will lead for further clarification of the question. First - when a boat sails into the wind (as the one in this question does), it is important to realize that (all these are idealized statements... real sailing is a lot more complicated): a) the force of the wind is approximately normal to the ... 0 A trajectory or flight path is the path that a moving object follows through space as a function of time. The object might be a projectile or a satellite, for example. It thus includes the meaning of orbit—the path of a planet, an asteroid or a comet as it travels around a central mass. Read more here : http://en.wikipedia.org/wiki/Trajectory I believe that ... 1 What a strange question. Is x(t) the distance measured along the track? Without defining x, there is nothing else to say. And if it is the distance, then it's the distance. Along the track. And the car is traveling along the track. So x is the distance. Which is what is asked. I suggest you read ... 0 When the slope is constant (left part of the drawing with the skier, flatness in the sled example) any possible acceleration is tangent to the surface. Thus, when the sled is speeding along perfectly level and flat ground, it experiences no acceleration. When the skier is going downhill, he experiences only acceleration along the slope. But, as already ... 2 Starting out with the general$$z := h + v~t - \frac {g}{2}~t^2,$$where$t \ge 0$is the duration since the stone had been released,$z[~t~]$is the remaining "height above ground" of the stone,$h$is the "heigt above ground" of ballon and stone at the release, and$v$is the vertical speed of the ballon at the release. Given$z[~\frac{2~u}{g}~] := ...

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