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## New answers tagged kinematics

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I assume, that you are helping your kid with homework. Actually the thata you have suggest that gravitional acceleration is different from 9.82 [m/s^2]. $v=u+gt$, so $100 \frac{m}{s} \neq 0 \frac{m}{s}+9.82\frac{m}{s^2}\cdot 10s=98.2\frac{m}{s}$ I will keep g = 10 [m/s^2] $h= \int_{t_0}^{t_1}\! v(t)\, \mathrm{d}t = \int_{t_0}^{t_1}\! u + gt\, ... -1 u = initial velocity (m/s): 0; v = final velocity (m/s): 100; g = acceleration of gravity (m/s^2): 9.82; t = time (s): 10; Use the equation v²-u² = 2aS; (100)² - (0)² = 2*9.82*h; 10000 = 19.64h; 10000/19.64 = h; h = 509.16m.; Height is 509.16m; 1 I agree with kevinsa5 that the variation in$a$is due to rounding errors, but I'd like to suggest a better way to analyse the data. Generally speaking, the best way to analyse data is to find a way to convert it to a straight line, then you can graph it and do a linear regression. In this case the way to procede is to note that if the acceleration is ... 0 The best thing to do is draw a graph of velocity versus time. Knowing the acceleration is constant, you draw the best straight line. The inclination will give you the acceleration. You can even draw the maximum inclined and least inclined line to determine your uncertainty. 2 You are right in that gravity did not change during data collection. You are a victim of uncertainty, which is a very important part of experimental physics. I'm sorry in advance for the "wall of text", and I hope that this clears up some confusion. The problem is that$1.50$may not be exactly$1.500000000...$. Because the numbers are provided rounded, ... 0 How can something move in a duration-less instant, when it has no time to move? According to elementary calculus, if the position of the arrow is given by the function$x(t)$, then $$x(t + dt) = x(t) + v(t)dt$$ This defines the instantaneous velocity: it is the ratio of the infinitesimal$^\dagger$change in position to the infinitesimal change in ... 7 Like all paradoxes, there is no contradiction here, just misuse of logic. How do you define velocity? If you say the distance traveled in an extended period of time, divided by that time well then of course there's no such thing as instantaneous velocity. Asking what something's instantaneous velocity is under this definition is logically equivalent ... 0 Centripetal acceleration = horizontal component of tension Weight = Vertical component of Tension$2$equations$2$variables 0 Well basically what you said is true, the maximum height, in this case, depends on the initial velocity and the angle$\theta$. So if you consider that the maximum height to be 2R, and by using the trajectory equation, while replacing$x_{y_{max}}=\frac{Total \;Distance}{2}=f(u)$, you'll get$u=f(R,\theta)$. Now I did the calculation and I got ... 2 You did not carry out your integration quite correctly. We have: $$a(t)= -B_0+B_1t$$ $$v(t) = \int\! a(t)dt=-B_0 t+\frac{1}{2}B_1t^2+C$$ Then plugging in conditions to solve for$C$we get: $$v(t_s)=0$$ $$0=-B_0t_s+\frac{1}{2}B_1t_s^2+C$$ $$C=B_0t_s-\frac{1}{2}B_1t_s^2$$ Now we can plug in$t=0$and solve for$v(0)$$... 1 The value of initial velocity will be different for different angles θ with the horizontal.. So I got this result.$$ u=(gR/(sinθcosθ-cos^{2}θ))^{1/2} $$or$$ u=(2gR/(sin2θ-2cos^{2}θ))^{1/2} $$or$$ u=(39.2/(sin2θ-2cos^{2}θ))^{1/2} $$This is my attempt for the solution(i have attached image): From A to B displacement is FB From C to B displacement is ... 1 HINT: Use H_{max}=\frac{u^2sin^2\theta}{2g} (max height) and R=\frac{u^2sin2\theta}g (Range). 1 I am not sure what you meant by: "I figured I could simply calculate the magnitude of the components since that will give me the distance" But the idea is use the kinematics equations for x and y: x(t)=x_{0}+v_{x0}t+1/2at^2 and y(t)=y_{0}+v_{y0}t+1/2at^2 These equations are derived from integrating the acceleration function ... 1 Suppose you have a satellite of mass m at a distance r. If we assume the satellite is small enough to behave as a point mass the moment of inertia of the satellite is:$$ I = m r^2 $$so its kinetic energy is:$$ E = \tfrac{1}{2} I w^2 = \tfrac{1}{2} m r^2 \omega^2 \tag{1}$$But for a body moving in a circle of radius r at an angular velocity ... 4 The equation comes from Newton's second law:$$ F = ma $$Galileo didn't know calculus (because Newton and Leibniz hadn't discovered it yet) so he couldn't derive the equation mathematically. Since we do know calculus we know that acceleration is the variation of velocity with time:$$ a = \frac{dv}{dt} And also the gravitational force F is equal to ... 0 You need to show some work. Draw a free body diagram. Then use Newton's laws with \sum F=0 (since you don't want the child to move i.e. accelerate) in both the x and y directions and solve the two equations for the unknowns. There are three forces: friction, weight (mg), and normal force. 0 It depends on whether or not you consider his angle at which he "jumped". Assuming he just ran straight off, we can simply just use our kinematic equations for both x and y. First let us do it for y because this will tell us how long he was in the air. First note note that he has no angle v_{y0}=0 \begin{align}y=&y_{0}+v_{y0}t+\tfrac{1}{2}at^2\\\ ... 1 The canoe has a relative velocity to the river which we simply call \vec{v}_{c/r} =( \dot{x}, \dot{y} ). The motion of the river relative to the earth is \vec{v}_{r/e} = ( 0.54, 0 ). The canoe relative to the earth is \vec{v}_{c/e} = ( 0.55\cos(-45^\circ), 0.55\sin(-45^\circ) ) . All together you have \vec{v}_{c/e} = \vec{v}_{c/r} + \vec{v}_{r/e} ... 0 You've done the "most difficult" which is to get as many independant equations as needed to solve for the unknowns. The first one flows from the fact that the vertical component of the swimmer's velocity is the only one to make him reach the end of the river, and the second one from the fact that both the current and the horizontal component of the swimmer's ... -1 [Edit: If you don't consider air friction and if you're not asked anything about the range of the projectile, the equations are the same as a vertical freefall]. If you are trying to find the velocity of the object at any given time, it is notv_{iy}$that you need to calculate since it is the initial velocity of the object at$t=0$. Step by step for a ... 1 This is correct, assuming constant acceleration, we have for this problem $$a = -\frac{v^2_i}{2d} = -\frac{(6.26)^2}{0.05} = -784\frac{m}{s^2}$$ First, I applaud you for asking the question. Too often, I have graded homework and tests where numbers were submitted for answers without any thought as to whether they were reasonable. However, this is a ... 1 Well, let's think about this: An object is traveling at 6.26 m/s during impact and travels only 0.025 meters before stopping. The force causing the object to decelerate needs to be extremely high. It's just like having a force being applied for a very short period of time, such as a bat hitting a baseball and the time of contact is extremely small, you ... 0 When you divide$25m$by$25s$you get$1 m/s$, not$1 s$What you should divide is the$100 m$circumference by$25 s$per revolution to get a speed of$4 m/s$. At the point W the direction is East. 0 Whenever a particle is constrained to move along a path, its velocity and acceleration vectors are decomposed along the tangent vector$\vec{e}$and the normal vector$\vec{n}as \boxed{ \begin{aligned} \vec{v} & = v \, \vec{e} \\ \vec{a} & = \dot{v} \, \vec{e} + \frac{v^2}{\rho}\, \vec{n} \end{aligned} } wherev$is the speed and$\rho$is ... 2 You're on the right track. A couple of notes: Those are actually total derivatives. You can think of$x(t)$and$y(t)$as functions of$t$alone you have two equations for two functions. You probably want to isolate them into two equations, each for one function Think about how you would solve this by elimination 0 The frame rate and the rotor frequency do not exactly match.the picture seems to be clear some times when the rotor reaches its set rpm. But after that the vibration starts with a different noise.the wobble increases and the distortion too.as Emilio pisanty said there is a vertical displacement due to vibration. 0 The way you pictured the problem, the particle is going to move forever in almost every case. Under the effect of conservative forces, the energy remains constant, so is going to move unless it reaches an stable point in which potential energy equals total initial energy. The curve you propose is an helix put horizontally, an the only way this can ... 5 One must distinguish between instantaneous velocity and average velocity. 1 Let's take a look just from the point of view of someone reading the problem. First of all, we can say "our car has motion", because it's changing its position each second. Ok. So, how is its motion? Well, it is moving in 1 dimention, it is a linear movement. Then, we can say "our car has linear motion". Also, we can see our car's velocity is changing ... 0 in v-t graph we see how velocity changes as time passes. if the graph is a straight line making acute angle with time-axis (x-axis), it means velocity is increasing at a constant rate. Now the problem here is, if i understood you correctly, is not thinking about it without using mathematics. In your question, the acceleration is constant and it is 10 meters ... 1 The car probably experiences a constant acceleration of$10{m/s^2}$. You can see from the chart that the velocity follows this as after every second the car is going$10{m/s^2}$faster. However, this is clearly not the whole picture. We do not know the acceleration at 1.5 seconds, or 1.55 or 3.14. We can get confirmation that our acceleration model works ... 0 You must consider the acceleration that is acting on the object. Let me give you an example, a ball falls from rest to the floor. This ball is 5m above the ground and the acceleration on the ball due to gravity is 10m/s^2. When the ball is let go, clearly the ball is not moving at 0m/s which is what it would be if there were no acceleration, it would just ... 9 The velocity is constantly increasing due to a constant acceleration. Exactly at 1 s the velocity is 10 m/s, but this does not mean that velocity was at 10 m/s in preceding second. In fact, given the distance 5 m moved in this second, the average velocity in this second was 5 m/s. And this should make sense to you, because in this first second the velocity ... 0 Hint : Speed of particle remains constant as magnetic field is always perpendicular to it. Also try to find a relation between the theta , D and h. You can also include R if you want. Try to draw the circle and its centre by using that the fact that normal at a point of a circle passes through the centre. Feel free to leave a comment below if you have any ... 1 First of all don't insert actual numbers until the end. It makes it much easier to keep track and check whether your units check out. This problem is easier if you invoke conservation of energy. Simply equate: At t = 0. -Potential gravitational energy. At the end. -Kinetic energy of the toolbox. -Dissipated energy due to friction. You will find that ... 0 Velocity is a vector, meaning it's direction must be taken into account. Acceleration is the change of velocity over time. On the crest the direction of the velocity changes continously, so there's an acceleration. In your hand drawing the sled passes over a horizontal terrain. Velocity does not change direction so there is no acceleration on the flat. 0 You are right that the difference between a crest and a trough is qualitatively just a mirroring. However, there is another subtle difference between the two cases. The skier is only on "level" ground for a single point along the path, while the sled on the hill is on level ground for an extended region. That is, your paths are not truly mirror images of one ... 3 You use for example python and mathplolib like in the following (slightly modified) example taken from http://faculty1.coloradocollege.edu/~sburns/toolbox/ODE_II.html. You need a working python/pylab installation to run it via python trajectory.py (if you save the file with this name) from the command line. The idea is to solve newtons equation of motion ... 0 Try adding the deceleration caused due to air resistance in the$x$-direction equation. Remember that the air resistance is velocity dependent, so you'll have something like this: $$F_{drag} = \frac{1}{2} \rho C_d A v^2$$ Which you would then put in the x-motion equation and integrate from time = zero to t 1 You are missing the fact that the truck is still moving forwards during its decelleration interval. 3 Relative position versus time is linear when relative velocity is constant i.e. when relative acceleration eqauls zero. In this case, both stones are acted upon by gravity only and hence their acceleration always equals$g$directed downwards. Therefore, relative position varies linearly for the entire duration of motion which stops when the stone thrown ... 0 Hint: The speed of the object number$i$($i=1$or$2$) will be $$v_i(t)=v_i(0)-gt.$$ 3 If you assume that Your body is a uniform, thin rigid rod. One end of the rod is pivoted (aka your feet) during the fall. Then one simply recalls that the angular velocity$\omega$of rotation of your body is related to the tangential velocity$v$of a point a distance$r\$ from the pivot by \begin{align} v = \omega r \end{align} Now, if you have height ...

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I think your analysis is good and correct. The statement you've quoted doesn't contradict what you've done, it probably suggests a different way to look at the problem. I think the statement simply states a method in mechanics- transferring a torque to another point (coordinate axes). In this case, from the centre of the wheel to the point of contact. ...

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