Kinematics Definition and 1000 Threads

if light pulse is to the right, but moving car's velocity is to the left, we would expect c-v, but we get c instead which is greater. but if light pulse is to the right, and moving car's velocity is also to the right, we would expect c+v, but we get c instead which is smaller. how could the same...

The solutions for (a) and (b) are pretty straightforward, which I got 13 kW and 225 kW each, but when I try to solve for (c), I get stuck with this: $$ \begin{align} a &= \frac{F}{m} \nonumber\\ &= \frac{F_\text{max}-f(v)}{m} \nonumber\\ &= \frac{7(30)+6(30)^2 -70v-6v^2}{1.000} \nonumber\\...

Imagine we kick the ball from point ##\text{A}## with horizontal speed ##u_x^{\text{initial, A}} = v \cos \alpha## and vertical speed ##u_y^{\text{initial, A}} = v \sin \alpha##. The gravitational acceleration is ##\vec g##, the x-axis points towards the wall, the y-axis points upwards. The...

I tried using the formula for acceleration in polar coordinates, but I don't know how to solve the differential equations. How do I solve them? Is there a simpler way to do the problem?

I tried to solve using trigonometry. Let angle between line CA ad BA be theta and angle between BA and AC be alpha. 30sin theta= 40 sin alpha 30 cos theta + 40 cos alpha + 2vt=50 I have no idea how to proceed after this.

So far I’ve dealt with scenarios in which the vehicle colliding with the rear of the other vehicle ‘joins’ the other vehicle, and the ‘two become one’, so there is one final velocity. Why have these two vehicles not ‘become one’? I’ll have a go at the maths first then have a go at answering...

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hello Take a 4 dof scara robot Calculate the x-y-z coordinates of the robot head , given the rotation angles of the 4 servo motors You can use the direct kinematics method Repeat this exercise a suitable number of times Train a neural net works of 3 neuron inputs (x,y,z) and 4 neurons outputs...

Is this absolute motion ?

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We know $$\vec{v}_{B/O} \equiv \vec{v}_{B/1}$$ $$v_{O/F} = v\hat{e}_t$$ $$a_{O/F} = \dot{v}\hat{e}_t + \frac{v^2}{R} \hat{e}_n$$\$$\omega_{1/F} = -\omega \hat{j}$$ $$\omega_{2/1} = \dot{\theta} \hat{i}$$Using velocity transfer relations, $$v_{B/F} = v_{B/1} + v_{O/F} + \omega_{1/F} \times OB$$...

With the given information , we can find ##y=h\cos\frac{\pi x}{L}## Since , the wheel has constant velocity ##v## , we can write the $$\vec{a}=0\hat{e_t}+\frac{v^2}{\rho+r}\hat{e_n}$$ where $$\rho=\frac{\left[1+\frac{h^2 \pi^2}{L^2}\sin\frac{\pi x}{L}^2\right]^{\frac{3}{2}}}{\mid...

So what i did was set up the cartesian axes ##e_1-e_2-e_3## and also the cylindrical ##e_r-e_\phi## The issue is how to calculate the cross products in the coriolis term without involving ##\phi##

This is for a math report that I'm supposed to write, which means I'm not supposed to use conservation of energy. This makes life much harder... so please bear with me. I am interested to see how you'd solve this purely kinematically though (if it can be solved that way). Please tell me if this...

Why can't I simply integrate a(x)=m*x with respect to x to determine the speed of a particle as a function of position v(x)=1/2*m*x^2+A?

Let's say the center of mass of the ring (wheel) moved a distance ##x##, tension in the thread connecting wheel and ##A## be ##T## and tension in the thread connecting ##A,B## be ##T'## : Due to no slip condition , ##a=R\alpha## where ##R## is the assumed radius of wheel Since ##P## is IAOR ...

Here I'm not worried about the solution as I got it. There are two ways I could come up with: Either finding the time it takes the cat to leap through the window, use that time to find what distance does the dog cross through the room Or: finding both times, for the dog and the cat to cross the...

I know That the velocity of each particle of the thread along the thread must be ##v_{0}## since the thread is inextensible. Now let's say the bead moves with some velocity ##v## For the particle attached to the bead to move with a velocity ##v_{0}## along the thread: ##v##cos##θ##= ##v_{0}##...

This is my solution below.

no clues

The time scale on which the change (such as a change in external parameters or a external parameters or an addition of heat) takes place is referred to as τ_exp. The relaxation time τ_relax, on the other hand, is the time that the system needs to return to a state of equilibrium after a sudden...

x=x0=vt v=v0+at x=x0+v0t+(1/2)at2 That (1/2) there does not make sense to me. It makes sense mathematically, if you integrate the velocity formula you get the (1/2) as a result of the integration rules and if you differentiate its needed to cancel out the 2 and arrive at the velocity...

Ive done this problem two different ways (sorry it’s messy) and keep getting hB = 1/9 hA, but my homework says it’s wrong. I’m guessing it’s because I assume that v0 is 0, but I’m not sure what other formulas or steps I need to take to either find or omit the variable. Are there any other...

Carlton writes on page 89: "The thrust of a marine propeller... may be expected to depend upon the following parameters: (a) The diameter (D) (b) the speed of advance (Va) (c) The rotational speed (n) (d) The density of the fluid (ρ) (e) The viscosity of the fluid (μ) (f) The static pressure of...

Attempt is attached, thanks for the help :)

For the first question I thought of using an energy balance, there is friction ##\Rightarrow \Delta E_m = -W_f##. Both at the start and at the end, the block has no velocity. Therefore ##E_{\text{initial}}= \frac 1 2 m_s v_{s,i}^2## and ##E_{\text{final}}= \frac 1 2 m_s v_{s,f}^2##. This means...

I'm trying to make a very basic physics engine. So far I've got a variety of small things worked out but I've been driving myself crazy trying to work out collisions. From one sense I get I can use momentum and impulse to determine the velocity of an object after a fully elastic collision (no...

For this problem, My solution to (a) is, We have constraint ##x + y = L##. There are many places we could define our (x,y) Cartesian coordinate system. However, the most easiest I think for the problem would be to attach a ##x^*## and ##y^*## coordinate system at the COM of ##m_1##. We define...

I managed to solve this by tilting the axes along the hill, and calculating the range, and then differentiating wrt ##\theta## (angle launched from hill) to get the answer. However, I recently came across the alternative solution below: The parabola it refers to represents the parabolic...

I get that: ##x(t) = A\cos(\omega t + \phi)## ##y(t) = A\sin(\omega t + \phi)## (from the above relevant equations). This agrees with the solution for part (a). However, the solution manual claims in part (b) that: In the case where ϕ1 = ϕ2 = 0 and A = B, the mass moves in a circle centered...

My guess was simply that as acceleration changes from the north to east direction, the total magnitude change of v is ##v \sqrt 2##. Acceleration is ##\mu g##, so time would be ##\frac {v \sqrt 2} {\mu g}##. This agrees with the textbook solution. What I do not understand is the trajectory...

This is the problem set. I am stuck from this point... If anyone could give me a hand I would really appreciate it. I know this is probably really simple, but I don't know any of this and have been trying my best with youtube, and other peoples posts. PS this is for high school

The distance covered by the first box is :s1max=v²/2|a|=v²/2μg where a=-μg by second newtons law Similarly S2max=(2v)²/2|a|=4v²/2μg It gas to be s1max+s2max≥S => v²/2a +4v²/2a ≥s => 5v²≥2aS =>v²≥ 2μgS/5=> v≥√(2μgs/5) But this is in the possible solution, am I wrong somewhere? I appreciate your help

My initial approach to this question was breaking the components of acceleration in the x and y axes and applying the three equations of motion to find the final velocity as well as the final position. As we were expected to find the net final velocity of the particle, I found the resultant of...

I'm using rigid body dynamics/kinematics in robotics stuff but I don't have a background in mechanics, I'm interested in understanding the kinematics of frame transformations for rigid bodies. Suppose we have two reference frames fixed on a rigid body, F_1 and F_2 and a transformation T which...

The answer should be 10 m/s^2 but I don't know how to solve it

The equation that connects final velocity with distance traveled is ##v_f^2 = v_i^2 + 2a \Delta y## Since the system starts from rest ##v_i = 0## and the above equation becomes. ##v_f^2 = 2a \Delta y## Since there is rotation in this system we need to connect ##a## to the rotation of the...

my attempt: i solved it all correct but i dont understand a few things mentioned above... 82.04 * v = 56 so i got v as 0.68 m/s which is correct but i dont understand the concept...

I think I have completed the exercise but since I have seldom used polar coordinates I would be grateful if someone would check out my work and tell me if I have done everything correctly. Thanks. My solution follows. Since ##\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1## it follows...

ay = -9.8. Vy = -9.8 + v0*Sin0. y = -4.9*t^2 + v0*t*Sin0. By using formula of y, my solution was -4.9(10s)^2 + (10m/s)(10s)Sin0. I assumed that y was equal to 0. Since -4.9 is m/s^2 the answer would be -490m + 100m*Sin0. Therefore I assumeed Sin0 = 490/100.

I tried using the formulas x=xi+vit and y=yi+voyt-1/2g(t^2) I assumed voy would be 0 and I almost arrive to the answer but idk how to get rid of the negative

this is how far i have come with my model, i am trying to first the most simple model, meaning no friction involved and then testing that against an actual stick falling by using tracking software. I am currently stuck as my model still has an acceleration in the y direction that i cannot seem...

A ball is thrown with an initial speed vi at an angle 𝜃i with the horizontal. The horizontal range of the ball is R, and the ball reaches a maximum height R/8. In terms of R and g, find the following. (a) the time interval during which the ball is in motion: Sroot(R/g) Correct (b) the ball's...

Why is it 10.92 seconds and not 10 Cars displacement = 30*t +1/2(0)t^2 police displacement = 0*t + 1/2(6)(t)^2 30t = 3t^2 t = 10 seconds ???

I propose to you a kinematics problem described by classical physics. Three space beacons A, B and C are 300,000,000 m (approximately one light second) apart. Beacon A emits a bright flash every three seconds. Beacons B and C respond instantly to the flash of Beacon A by emitting...

idk

the answer to this question uses the above formula with the tangent function and solves for the initial velocity, i used the equation (v.sinθ^2) = (v.sinθ)^2 - 2gΔy, setting final velocity equal to zero and solving for initial velocity. this kinematic equation gives a different answer. can...

[FONT=times new roman]1. [FONT=times new roman]The first equation between velocity ##v## and time ##t## can be derived using the graph I have drawn for the purpose as shown on the right. Since acceleration ##a_0## is a constant, the graph of ##v-t## is a straight line. The slope of the line is...