Kinematic Definition and 395 Threads

Hi, so I first converted the speeds to meters per second (m/s), and then proceed to finding if they will clash using 2ad = v^2 - v0^2 assuming v0^2 is zero to find their final position. Red train speed: 20 m/s Green train speed: 40 m/s The stopping distance d for red train is 200 m and the...

Wild question. Even Chat GPT struggling. What I first attempted was to find the velocity by which the projectile with v1 falls down to the ground with. This part was easy. Kinetic energy gained = gravitational potential energy lost Let's define our coordinate frame as up vectors being positive...

A-A' (mv1^2)/2-(mv0^2)/2 = A A = -AG + AT - AF AG = mgh = 1.47 J AT = k/2*h0^2 = 0.0125 J AF = fmg*cos(60)*h0 = 0.0735 J A = -AG*cos(30) + AT - AF = -1.334 A-C (mv1^2)/2-(mv0^2)/2 = A A = AG C NC + Gcos(45) - Φn = 0

##\vec n_2## points to the diagonal top left. The velocity v is a function of t. So for example ##2t^2## and a = dv/dt. Putting the velocity vector into ##\vec n_1## and ##\vec n_2## terms. $$\vec v = v sin\theta \vec n_1 + v cos\theta \vec n_2$$ $$\vec v = -v \vec i$$ $$\vec a = \frac{d\vec...

I am super stumped at this question, the answer key is telling me 6.78 downwards, i think I'm reading and observing the question wrong. Isn't final velocity 0 so why can't i do 0 = v0 -9.8(0.25) -2.45 = v0 But when I use the second the equation -2 (displacement of door) = v0(0.25) + 1/2...

This is what I've done so far: 54 km/h = m/s 72 km/h = m/s a = 2,0 m/s^2 A = d_1 = v * t = 15 * 10 = 150 m B = d_2 = v * t = 20 * 10 = 200 m d_3 = d_1 - d_2 = 200 - 150 = 50 m Don‘t know how to continue to solve the problem.

Cycloid is a curve which can be defined as a trajectory of a point marked on the rim of a rolling wheel or radius R. Determine the curvature radius of such curve at its highest point. what you need to do is to equate 4v²/r with v²/R and to get that r=4R but i dont understand why the answer...

Hi, I was wondering if you could help me with a job at the university that consists of the kinematic modeling of mechanism number 3 in the image attached to this message. I have to set up the constraint equations and then solve the position, velocity and acceleration problem. For now I would...

For ## d = v_it + 0.5at^2##, can we solve for t without using the quadratic formula? Many thanks!

How to find the collision number if the moving bullet hits a few wooden blocks and every collision takes 10 percent of its speed. In which block will the bullet stay?

We usually have an initial time and then find an equation for the variable final time. Can we derive a formula to calculate position with final time and variable initial time. ##v = v_i + a(t_f - t_i)## ##dx = v_idt + at_fdt - atdt## integrating ##x_f - x_i = v_i(t_f - t_i) + at_f(t_f - t) -...

The attached pdf is my solution. Thank you!

This is not really a homework problem (it could be made to be though). I kind of made it up, inspired by a youtube math challenge problem involving parabolas, a water fountain where A = 1, R = 3, and H = 3. The solution given (h = 9/4) was based off simple math utilizing vertex form of a...

So what I have done is that on question 7 I know he is accelerating to constant my knowns are that it is from initial and acceleration is given so I have that for my first phase the second phase is that Your vfinal is your new initial for it and acceleration is the constant but you don't know...

I'm not sure where to start, when I tired using integration of the initial equation to get pos(t)=-.65t^2 i + .13t^2 j + 14ti +13tj but after separating each component, i and j, and setting j equal to zero I got 0 or -100 seconds which doesn't seem like a reasonable answer.

Givens: Vyi=12.5 m/s Vyf=-12.5 m/s (at the same horizontal level) ay=-9.81 m/s^2 Δy= zero m (as the displacement on the y-axis, when the projectile reaches the same horizontal level, is zero m) Δt=? When I use Δy=[(vyi+vyf)/2]*Δt I get the time as undefined. Δt= 2Δy/(vyi+vyf) = 2*0 m/(12.5...

In a recent thread, I said that if there was interest, I would post in a separate thread the calculations for the kinematic decomposition of the congruence of worldlines describing the rod in the "rod and hole" relativity paradox discussed in that thread. Since there was interest, I am posting...

Would we assume that the deceleration of both instance are the same?

Hello! I have done this problem : vf^2 = (4.0x10^5)^2 + 2(6.0x10^12)(5x10^-3) so vf= sqrt((4.0x10^5)^2 + 2(6.0x10^12)(5x10^-3)) I get vf = 4.7 x 10^5 m/s However, the textbook solutions says vf = 8.7x10^5 m/s. Where did I go wrong? Thank you for any help! :)

i was doing some problem and i have a hard time figuring out when will the y be positive and negative and same with the gravity idk if it -9.8 or 9.8

I've written in that only the first statement was False, however my classmates aren't getting the same answer? Please help clarify...

##V_x = V_0 cos \theta ## ##x = V_0 cos \theta t## ##V_y = V_0 cos \theta ## ##y = V_0 cos \theta t## ##F_x = m\ddot{x}## ##-mgsin \phi = m\ddot{x}## ##\dot{x} = -gtsin\phi + V_x## ##x = -\frac{1}{2} gt^2 sin \phi + V_x t## ##x = -\frac{1}{2} gt^2 sin \phi + v_0 cos\theta t## ##F_y =...

Can anyone help me with a kinematic analysis for the mechanism attached below? I need the outline of the mechanism, its notes and the necessary formulas to find out the transmission ratio and the rest of the values.

Vƒ = velocity final Vi = velocity initial a=acceleration t=time 0.5 = ½ ^2 = squared - = minus d = displacement Equation d = Vƒt - 0.5at^2

I have a question : If we consider the change in g due to distance from the Earth core; then y=distance from earth’s core t=time G=gravitation constant M=Earth’s mass k=GM $$y^2(t)=\frac{k}{y(t)^2}$$ If we consider air resistive force as proportional to speed squared, then: m=falling object...

Hi guys, given the "blacker" to be the cabin under consideration, I firstly wrote its weight force; then, my confusion started when drawing the force applied on the cabin by the structure(##F_{r}##). I concluded it must have been both counter-acting the weight, and acting as a centripetal...

please I need to clarify this question thanks sincerely Luis

Details of Question: ds/dt= v which becomes ds=v dt, where s=displacement, t =time, and v=velocity Then we can integrate both sides of this equation, and do a little algebra, and turn the above equation into: s − s0 = v0t + ½at2 My main question is about the integration of...

F = qE ma = (2*10^-6) * (λ / (2pi*r*ε0) ) ma = (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) => I am not certain what to put for r ( But I sub in 4 because dist is 4) a = ( (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) )/ 0.1 a = 0.35950 v^2 = U^2 + 2 a s v = 0 u^2 = -2 a s => Can't sqrt negative so...

So, what I did was suppose the mass of ramp is $ M_r$ and let velocity at B of block be v, then, after inellastic collsion both bodies v' velocity at B , $$M\vec{v}= M_r \vec{v'}+ M \vec{v'}$$ or, $$ \frac{M}{M +M_r} \vec{v}= \vec{v'}$$ Now, Suppose I take the limit as mass of ramp goes to...

Summary:: Just a simple 3d rigid dynamics question which I am trying to solve by placing coordinat system differently from original solution.Everything looks ok but results are different. Mod note: Post moved from technical section. Thats my question.As you see coordinate system was located...

The first thing I did, was to find the equations for player A (p) and ball's (b) path (for each i and j component I used the equation I wrote in the relevant equations) and then I found the derivative of both equations so I could have the velocity: $$\vec{r}_p(t)=(6t^2+3t)\hat{i}+20\hat{j}...

stuck on this question

the red line is the initial velocity, the grey parabola is the path of the projectile. hi there...I'm kinda stuck at the part b of this problem. I can do part a with no problem. can anybody explain to me how to do the differentiation needed to solve part b?? by explain I mean explain the...

My attempt at a solution is to start off first denoting V_a to be the automobile an V_e to be the economy version. Same goes with l_a and l_e. To try and relate the two I have tried: V_a I_a = V_l L_e, however I am really not sure how they got the square root. The answer is: v = V sqrt(l/L)...

I've been attempting to solve this problem for three days now. I have thrown away my old attempts (like, scrumpled up into the bin), but my old attempts involved: Trying to set up simultaeneous equations relating the journeys between EH and FG to find the deceleration, but the reason why this...

Homework Statement: Three particles A, B, and C, each of mass 𝑚, lie at rest in that order in a straight line on a smooth horizontal table. The particle A is then projected directly towards B with speed 𝑢. The masses of A, B, and C are now 𝑚, 2𝑚 and 3𝑚 respectively: Again find the fraction of...

I just ran this pretty quick at work. But this is the general outline. Sorry for the slop, it will get better with time. Thanks in advance and any additional info can be supplied. The material is from the Khan free course. ..... A student is fed up with doing her kinematic formula homework...

Part A) So, I set the two positions equations of A and B equal to each other since the position has to be the same. A: 0 + 0 + 1/2(2)t^2 = 50 + 0 + 1/2(4)t^2 :B I know I have to solve for time t, but there's no way to solve it with both sides having t^2 so I am not sure which variables I got...

What should I do? Because I have two possibilities. I have ##0=5+at## so ##-5/t =a##. But then I can also say that the acceleration is a negative because it is stopping, so I can write it like ##0=5-kt## and then ##5/t =k##

The first doubt that comes to my mind is "I have to determine the acceleration with respect to what?", because the problem doesn't tell. Then, I have some problems when having to plug the data in the formula of acceleration. ##\vec a_B=0## because the origin isn't accelerated, ##\vec{\dot...

So my problem isn't actually finding the components, but knowing if the initial approach I took is correct. So what I did was: At first I found that at the same instant, ##x_{B/A}=10500 m## so then I wrote the equation of motion for plane B respect to A: so $$\vec a_{B/O}- \vec a_{A/O}=\vec...

So what I did was at first consider the case the kid is below the branch, so that x=0,t=0, then I thought that the length L of the rope should be ##L=2h## because we know the radius from the branch to the kid is just ##x^2+y^2=r^2## and when x=0, y=h. So then I wrote the motion equations for the...

So I know that ##a_t = \frac{dv}{dt}=-ks## and ##\frac{dv}{dt}=v\frac{dv}{ds}## then: $$v dv=-ks ds \rightarrow (v(s))^2=-ks^2+c$$ and using my initial conditions it follows that: $$(3.6)^2=c \approx 13$$ and $$(1.8)^2=13-5.4k \rightarrow k=1.8 \rightarrow (v(s))^2=13-1.8s$$ What bothers me is...

I tried to workout the problem but I find motion in different coordinates systems a bit weird at the moment, so only thing I could do is realize that the x component of ##\vec r(t)## is: $$vt +x_0$$ but for simplicity we will use the initial condition ##x_0=0## so that ##t_0## is the moment the...

So ##T+U=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})-mgy=constant##. If I derive this with respect to ##t## $$\dot{x}\ddot{x}+\dot{y}\ddot{y}-g\dot{y}=0$$ Then I use ##\dot{y}=\dot{x}\frac{dy}{dx},\ddot{y}=\ddot{x}\frac{dy}{dx}+\dot{x}^{2}\frac{d^{2}y}{dx^{2}}## to get...

I've calculated the potential energy at the top of the halfpipe, before the boarder drops in: PE = 39.5 kg * 9.8 m/s^2 * 3.66 m = 1416 J Since the boarder would have no potential energy and all kinetic energy at the bottom of the halfpipe, KE = 1/2mv^2 = 1416 J 1/2 (39.5 kg) (v^2) = 1416 J So...