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## New answers tagged kinematics

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Your equations: $\Delta d_y=v_1$$_y$$\frac {\Delta d_x}{v_x}+\frac {1}{2}a(\frac {\Delta d_x}{v_x})^2$ $\Delta d_y=(13.7m/s)(\frac {100.0m}{37.6m/s})+\frac {1}{2}(-9.8m/s^2)(\frac {100.0m}{37.6m/s})^2$ The question of whether you should put a v_1$$_y really depends on your choice of initial condition. Take note that the cannonball starts at cliff height ... 0 It's not clear to me how you have got the values of v_x and v_y you used in the final part of the question and given the info in the question I don't think they are correct. I think you are missing a critical point at the start. You can easily calculate the time of flight as 10s as you know the boat speed and how far it traveled in that time. You can ... 0 Use conservation of energy.Kinetic energy of vertical component is converted to potential energy.Use this to find the vertical velocity at launch.Also tan theta =Vy/Vx.from this you can find the horizontal velocity.for final velocity, add them vectorially. 5$$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=\frac{dv}{dx}v$$So the differential equation with x as the independent variable becomes$$v(v'+1)=1$$0 By conservation of energy, if initial velocity is v then the velocity at an angle a is given by u=\sqrt{v^2 -2gR(1-\cos a)}. Now let the object move through a distance ds. ds=R\cdot da (da=small angle subtended by ds at centre). Use dt=\frac{ds}{v}. Integrate RHS from angle 0 to a and similarly integrate LHS from time 0 to t. Now ... -1$$ Range=H_m \space_a\space_x  u^2sin2\theta/g=u^2sin^2\theta/2g  tan^-\space^1\space4=\theta $$WIKIPEDIA CLICK HERE TO SEE FORMULAE 2 What I have done is to add the 569\;\mathrm{N} vector with μ_kn and finding the horizontal component of that resultant vector. Something's wrong here. The resultant force vector should not have any other components than the horizontal one - otherwise the object should be moving vertically also, which it doesn't. In the sentence here it seems that ... 0 So we have a z axis pointing upwards and a object in free fall from z=z_0 to z=0. The equation of motion in free fall is -ma=mg or a=-g. \large{a=\frac{dv}{dt}}. dv=-gdt and integrated we get: v=\int_0^t(-g)dt=-gt. v=\frac{dz}{dt}, so dz=vdt, dz=-gtdt and integrated we get: z=z_0-\frac{gt^2}{2}. For z=0, ... 2 So you know the radius of the two balls, call them r for the small ball and R for the bigger ball. However, like the commenter Bernhard says, the terminal velocity of each ball will be different. The terminal velocity is:$$ v_{t}\propto{\sqrt{\frac{m}{\rho}}} $$Now you can begin to plug into the equation and you should be able to see that the ... 1 I read your question. The answer is that the stone will have upward velocity v and downward acceleration g. Now I will explain my answer by correcting your attempt. I have read your question thoroughly. You have made a few mistakes in your calculation. \vec v_{s,b}= v and \vec v_b= 0 (I am not too sure of this, that is why the question) Which gives ... -2 With respect to the balloon the stone will have zero velocity and an acceleration of g + a downwards.Especially if the balloon is moving up. 1 Factor the first term:$$\Delta x= \dfrac{v(t)-v_0}{2}\left[\dfrac{v(t)-v_0}{a}\right] + v_0\left[\dfrac{v(t)-v_0}{a}\right]= \left(\dfrac{v(t)-v_0+2v_0}{2}\right)\left[\dfrac{v(t)-v_0}{a}\right]= \left(\dfrac{v(t)+ v_0}{2}\right)\left[\dfrac{v(t)-v_0}{a}\right]\Delta x= \dfrac{v(t)^2-v_0^2}{2a}$$-1 Actually, due to lack of friction, you could say m_1 would remain right there and M would move underneath it. But, you could say it the other way round. M is still and m_1 moves to the left with acceleration \frac{m_1a_M-T}{m_1}. Now, m_2 would have an acceleration of \frac{T-m_2g}{m_2}. Now, both these accelerations are same. Solve for ... 0 They both must travel the same displacement in order to collide with each other. Try using the kinematic equation s=ut+\frac{1}{2}at^2. Remember that as Socre said, one will have travelled the displacement in x seconds, while other in (x-1) seconds as it was thrown 1 second after and that the displacement of the ball 1 = the displacement of the ball 2 ... 0 Your equations for tension are all correct. You have three equations and three unknowns T, a, and a_M so you can solve for that system. If you'd like to know the normal force between M and m_2 you can calculate that as:$$N=m_2\,a_M$$1 The shortest leg gets the most weight and the longest leg gets the least weight. With a triangle ABC with the center of gravity positioned over point O the weight's that each leg will hold would be as follows:$$W_a=W\frac{BO\times BC}{BA\times BC}W_b=W\frac{AO\times AC}{AB\times AC}W_c=W\frac{AO\times AB}{AC\times AB}$$Where W is the total ... 2 At school physics level the vast majority of systems you need to analyse are described by one of the SUVAT equations. Sadly the Wikipedia article on these got deleted, but Googling SUVAT equations will find you lots of related articles. Searching this site will find lots of related questions. The equations are all derived from Newton's second law as ... 0 You're very close. Instead of changing reference frame so that an end is stationary. Change reference frame so that the center is stationary. The rod will then spin about the center. Your equation for \dot \theta is correct (though you're using different coordinate systems for your two velocities so the signs are strange)$$\dot \theta=\frac{\Delta ... 0 Straight Newton's Laws plus kinematics solution An Atwood's machine develops an acceleration of \begin{align*} a &= \frac{m_2 - m_1}{m_2 + m_1} g \,, \end{align*} in the direction of the heavy mass (herem_2$). Both objects have this acceleration though in opposite directions, and (assuming the rising one doesn't hit the pulley) it persists over ... 0 Rapidity has a number of related advantages, as you and the accepted answer have pointed out. It also has a geometric interpretation. See this SE answer (but note that that answer answers a somewhat different question). 0 Rapidity is especially meaningful in the context of Galileo's relativity principle. That is, if you ride an inertial frame, and use the same physics to impart the same impulse to yourself to accelerate to a new, colinear moving frame, then this "standard physical manoeuvre" will always lead to the same rapidity change relative to your initial frame. As a ... 0 You can use Galileo's principle of odd numbers which sates that: if a body is undergoing motion under constant acceleration(9.8m/s/s here)and its initial velocity was zero then the distance covered by it in equal intervals of time are in the ratio 1:3:5:7...(that is odd numbers); So in 1st second distance covered =H Assume in the 2nd second distance ... 0 Search for Kinematic Equations. You should find something that resembles$y = y_0 + v_y t + .5gt^2$where g = -9.8$m/s^2$If the particle starts falling from rest ($v_y$= 0) at a height$y_0$=60 meters then after 1 second the particle will fall $$y=60 - 4.9(1)^2 = 55.1 m$$ The difference in height is$y - y_0 = -4.9m = H$. The negative sign indicates ... -1 From looking at the chart I would have to agree with everyone who said yes. But not roll but instead it would flip. Do to there being no friction stopping the cart from doing so and the hanging wait being heavier then the resting weight. If you add friction and weight of the cart you may come up with a variable that would fit the theory that the cart ... 2 Yes, the cart will move, due to the force applied by the string to the pulley. To solve, calculate the string tension while the weights are moving, and then note that the pulley has to provide an opposing force in order to change the string's direction. The reaction to that force acts upon the cart, accelerating it. Momentum is conserved, because the ... 0 The no math answer to this one is to realize that acceleration is the rate of change of velocity. At the top most point, the velocity is indeed zero. However, it is changing momentarily after that. If the acceleration was zero, the ball would have had no change in velocity and would have stayed up in the air forever. 0 You are correct, in that the velocity is zero, so its direction doesn't mean anything, but just because the velocity is zero doesn't mean the acceleration is zero. And it's not zero, it's -9.8m/s/s, as you acknowledge, so the direction of acceleration is meaningful. 2 You throw the ball upwards with velocity$v$and it returns to your hand with velocity$-v$. Let's draw a graph showing the velocity as a function of time: Acceleration is defined as: $$a = \frac{dv}{dt}$$ so it is the gradient of the line in this graph. The velocity-time line is straight so the gradient is constant which means the acceleration is ... 0 I think you are subconsciously mixing up velocity with acceleration. Let me give you an example. Imagine these are the measured speeds of a particle thrown vertically into the air at different times: time, speed 0s, 50m/s 1s, 40m/s 2s, 30m/s 3s, 20m/s 4s, 10m/s 5s, 0m/s 6s, -10m/s 7s, -20m/s 8s, -30m/s 9s, -40m/s 10s, -50m/s <-- it hits ... 1 When you shoot the ball upwardly, gravity acts on it with a force$mg$where$m$is the mass of the ball and$g=9.81 ms^{-2}$the Earth's gravitational acceleration. If the initial upward velocity was$v_0$then the instantaneous velocity$v$is given by:$v=v_0-gt$, so after some time$t=\frac{v_0}{g}$the balls's velocity becomes$v=0$. However, we know ... 0 At the topmost point, the velocity vector is a null vector whereas the acceleration vector has constant magnitude$-9.8\,\mathrm{m/s^2}$and constant direction downwards i.e. towards the centre of earth. 1 Place the car's centre of gravity at the origin of a Cartesian coordinate system$x,y$, as shown above. Now determine the balance of forces along the$x$axis. Assume the car will slide (up or down), as there's no friction. The equation of motion is:$ma_x=mg\sin\alpha-F_c\cos\alpha$(Eq.1).$F_c=\frac{mv^2}{R}$with$v$the velocity of the car and$R$... 0 The ball is rolling along positive x direction with velocity v. The points of the ball directly below the center are rotating around the center of the ball in the -x direction with instantaneous velocity u equal to the distance from the center of the ball times the angular velocity of the ball. Where v = u you have the condition you are looking for. Truly ... 2 Physicists tend to be a bit casual about sign conventions when it seems to be obvious. So let's attempt to be completely rigourous. The key step is getting the flight time$t$since the range is just$v\cos\theta\, t$. We do this using the SUVAT equation: $$v = u + at$$ We'll use the usual conventions that up and right are positive, so$v_y$and$v_x$... 3 You are looking at a specific application of a more general formula $$v_q^2-v_{oq}^2=2a_q(q-q_o),$$ where$q$is the coordinate direction, the$v$terms are velocity components along the$q$axis,$a_q$is the constant acceleration component along the$q$axis, and$q$and$q_o$are the positions along the$q$axis, which match, respectively with$v_q$... 0 It doesn't say that. If the positive axis is "up" as you put it, when the initial velocity is "downwards" that means that h is always negative, meaning the -2gh term goes positive. 1 Setting $$\theta := \theta_1 + \theta_2$$, the momentum of the Higgs boson (candidate) with respect to the lab $$\| \textbf p_{lab}[~H~] \| = \| \textbf p_{lab}[~\gamma_1~] \| ~\text{Cos}[~\theta_1~] + \| \textbf p_{lab}[~\gamma_2~] \| ~\text{Cos}[~\theta_2~] = (E_{lab}[~\gamma_1~] ~ \text{Cos}[~\theta_1~] + E_{lab}[~\gamma_2~] ~ ... 0 A couple of other things that can be considered: - As the car's speed approaches zero, static friction will take over from dynamic friction in the brakes, which will suddenly increase the deceleration if constant pressure is maintained on the brake pedal. This effect can be avoided by reducing the pressure at the last moment. - The effect of the springs ... 0 Hint: Begin by showing that the general solution of the linear first-order differential equation$$ a=\frac{dv}{dt}=-0.5v $$with constant coefficients is$$ v(t)=v(0)e^{-0.5t}. $$4 It isn't clear from your question exactly what you are integrating and how, but this is the way to tackle problems like this. You know that:$$ \frac{dv}{dt} = -kv $$The way to solve equations like this one is to rearrange it by dividing both sides by v and multiplying both sides by dt to get:$$ \frac{1}{v}dv = -k\,dt $$Now we can integrate both ... 2 Fast things appear blurry to your eyes, and the distance from a point on the paper to your pencil is fixed. The distance from an ink dot on the paper, to the tip of your pencil where you press down, cannot change. Call this distance d. The only positions the point is allowed to be at when you spin the page around, are positions that satisfy the equation ... 0 Proceed through the following steps: Differentiate your equation with respect to t to get some relation between \frac{dx}{dt} and t. Name this equation (2) and the given one (1). Now solve the first equation for x=0 and you will get a real positive value of t. Put this value of t in equation (2) to get the answer to your first question. Now solve ... 0 You've confused velocity and speed. Speed is independent of direction, and is therefor always positive. For a 2-dimensional analysis, you have to consider the two different components of the velocity, Vx and Vy, but the speed S is$$S=\sqrt{V_x^2 + V_y^2}$$Let's take a version of your example. An object travels along the x-axis at 5 m/s in the positive ... 0 Recall that velocity is given by$$v \equiv \frac{\mathrm{d}x}{\mathrm{d}t}$$As a result, to determine your particle velocity at a given time, simple differentiate your position equation, and then solve for your particular condition (x=0) 0 we first define velocity with the following equation$$v=\int{a}dt$$Differentiating the equation would yield acceleration which is incorrect based on what the question is asking You stated your equation for velocity is velocity$$v=f(t)=2cm/s^3t^2+5cm/s$$Therefor by simple substitution we just sub v=f(4)$$f(t)=2\times(4)^2+5=37cm/s$$Verifying the ... 0 When you have a function f(t) that expresses a quantity (say the velocity v), then you just evaluate that function at t in order to get the function (velocity) at t. However, if I give you the position at t, x(t), and I ask you for the velocity, then you first have to recall that velocity is the derivative of position with time:$$v = ... 7 I know how , in the physical sense - $$\frac {dv}{dt} = a$$ But, even after thinking a lot, I am not able to see the fault in this - $$\frac {dv}{dt} = \frac {d(st^{-1})}{dt} = \frac {sd(t^{-1})}{dt} = s*(-1)*t^{-2} = \frac {-s}{t^2} = \frac > {-v}{t} = -a$$ I know something is terribly wrong here but I'm just not able to ... 2 Yes. This approach is definitely viable. Coincidentally, equating the time-derivative of vertical position to 0 is the same thing as maximising the height mathematically (concept of maxima and minima). Another approach is to use the conservation of energy. Assume the vertical component of velocity to be$v_y$. Solve it by equating difference of potential ... 1 Let's approach this slightly differently; I am sure you would agree that: $$v a = a v$$ This is the equivalent of your equation after being multiplied by$v$where$v=\frac{dx}{dt}$and$a=\frac{dv}{dt}\$. Now integrate this with respect to time: $$\int_{t_1}^{t_2} v a dt = \int_{t_1}^{t_2} a v dt$$ From the definition of the derivatives we know that ...

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What is said in the above comments is true. The kinetic energy stays the same because the absence of a net force implies that the only non-vanishing contribution to the energy of the body is kinetic energy. Since the total energy is conserved, the result follows.

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