Kinematics Definition and 1000 Threads

Show that a point with acceleration given by: a=c*((dr/dt)×r)/|r|3 where c is a constant, moves on the surface of a cone. This is jut an example to illustrate my doubt. I don't know how to obtain the tracjectory given only the acceleration in this format. I realized that if i can show that...

Im for some reason getting 1.58 s for time. I found 1.75 m as the head start distance and then I do d=d so: 2.45t^2 = 1.75t^2 +1.75 but the answer for "a" should be 6.45s...

I solved this and got 27.3 seconds so basically what I did was found the headtsart of the slower car (15.6 m/s times 6.65 seconds) but the answer given is 33.9 seconds and it uses the second (faster cars speed) to find the head start distance. Which answer is correct and why? Thanks so much

Pic 1: Pic 2:

i calculated the time difference between each marker, and using the first equation, i solved for V1. i then used this V1 as V2, time as the first marker (3.75) and used the second equation to solve for V1. is this method correct? can somebody please verify?

Hi! I can't solve this. Please someone give me a hint and help? I'm unsure what equation to use. Thank you!🙏🙏🙏 An artillery crew demonstrates its skill by firing a shell at an angle of 49 deg and then lowering the gun barrel and firing a second shell at a smaller angle of 20 deg in such a way...

1. $$x_{A}(t)=\begin{cases}\frac{1}{2}a_A t^2 & 0\leq t\leq 2.8\\ \frac{1}{2}a_A 2.8^2 +v_A t & t>2.8\end{cases}=\begin{cases}\frac{1}{2}a_A t^2 & 0\leq t\leq 2.8\\ \frac{1}{2}a_A 2.8^2 +a_A\cdot 2.8\cdot t & t>2.8\end{cases}$$ so ##x_A(5)=40## implies ##a_A=\frac{40}{17.92}m/s##. Similarly for...

Does the position of the origin for the body’s rotating coordinate frame 1) stay fixed to the moving body or 2) does it stay fixed to the inertial frame, yet still able to rotate as the body rotates with the only restriction that it cannot translate with the body i.e. only affixed at the...

0=6-4.9(t^2) 0=-27,7t+Xo From the first equation, T= 1.10s From the second equation, (Xo/27,7)=t 0=6-4.9((Xo/27,7)^2) We solve for Xo = 30,65 Now, we have X=-27,7t+30,65 If t=1,10 Then , -27,7(1,10)+30,65=0,18 which is not equal to 0. Which means the stone hits the ground first. Is my answer...

Hello! I have a somewhat complicated robot project that I've simplified here to hopefully get a quick answer. I want to build a robot with two omni-wheels that looks like the following (not drawn to scale), where the black is the omni wheels at 45 degree angles and the blue is the robot body...

I was taught that kinematics is the study of local motion without references to its causes (like forces, fields, etc.). That seems useful to say but perhaps it is not really that helpful. What are the kinds of kinematics? I can think of three kinds: classical kinematics (from mean speed to...

I had deduced that B,E,H are the places where acceleration will be zero, but when I read the solutions it showed that K also has a = 0. It said it had maximum slope and then said a = 0. But I couldn't understand why? Please help.

We have direct mass models of stellar motion inside of galaxies (basically, find a happy medium between a sphere and a disc). And of course dark matter corrections to explain the deviation between model and observation. Is there any pattern at all to the motion of extragalactic stars passing...

The way I tried to solve it is by using following reasoning. The inclined part of the string will each have a speed of U m/s. So the point P just above mass M where the inclined strings meet is being moved up by each inclined string at ##U \cos {\theta} ##. Therefore, the speed of the point P...

The point of confusion is deciding the direction each persons sets out in i.e. velocity direction of each person. Knowing this will probably help in getting the solution. At t=0, I can say that velocity of each person is as shown in diagram below.

By considering a vector triangle at any point on its circular path, at angle theta from the x -axis, We can obtain that: (rw)^2 + (kV)^2 - 2(rw)(kV)cos(90 + theta) = V^2 This can be rearranged to get: (r thetadot)^2 + (kV)^2 + 2 (r* thetadot)(kV)sin theta = V^2. I know that I must somehow...

I have the following problem: "A box with a mass of 10 kg is accelerated by 10 m/s/s over a distance of 10 meters. What is the kinetic energy of the box (assuming no friction)" Using work... Work = K.E. F * D = K.E. m*a*D = K.E. = 10 kg * 10 m/s/s * 10 m = 1000 Joules However, using...

Help please

t= v/a =25.8 m/s / -2.5 m/s^2 =10.1 s therefore, d= vi * t + 1/2at^2 = (25.8)(10.1) +1/2(-2.55)(10.1)^2 =131 m I got the first one right I think, but I don't know how to do b

I'm trying to simulate the physics of a Toyota Camry during acceleration with a time granularity of 100ms. My simulated conditions are as follows: m = 1590 kg v = 17 m/s a = 1.5 m/s2 η (transmission efficiency) = 0.85 rwheel = 0.35 m Fdrag = 100 N Ffriction = 260 N Faccel = 1590 kg × 1.5 m/s2 =...

(a) Using COE, $$mgh = 0.5mv^2 + 0.5I\omega^2$$ I solved it, where $$\omega = 112 rad/s$$ (b) This is the part where I have question or problem. I saw my course mate working and he start of with finding centripetal acceleration. $$a_c = \frac{v^2}{r} = \frac{(r_0\omega)^2}{R_0}$$ Why isn't it...

I rearranged the displacement formula to d2 = d + d1. I used cosine law to solve for d2 since the triangle is not right-angled but I am not getting the correct answer or angle for d2. The angle I used in cosine law (based on the diagram) was 32+12+90 = 134. d = v(t) = 130(3) = 390 km/h [N 32 E]...

Hello, Perhaps I am overthinking this, but I am trying to find the acceleration of the 'pulse rate' in the below plot. Every so many seconds there is a pulse (you can think of this as x displacement). The initial rate between pulses is 5 pulses/sec (you can think of this as velocity). I am...

I don't know why I'm so puzzled by this problem; it's only one star. So first, I drew a picture of A and B in the ground frame. Then I drew B and the ground in A's frame. I then used the velocity addition formula to obtain the velocities of both B and the ground relative to A. $$\frac{v_{B}^{A}...

I’ve read Beiser’s Modern Physics Chapter 1 and I am able to grasp the general theories but not so much when applying it to problems. I major in Chemistry and would really appreciate any head start/help/suggestions. For number 1 I thought of using L = L0 sqrt 1-v^2/c^2 but can’t seem to find the...

I found vlog=1.9/t and vb-v=1.15/t and vb+v=d+1.15/60.8 don't know what to do now

Do i use the equation above mentioned to solve it? If i do, am i correct to say that Vx = 90/3.6 cos 30 and Vy = 90/3.6 sin30?

m1 + m2 = 8 COE 0.5(m1)(u1)^2 + (m1)(g)(30) + 0.5(m2)(u2)^2 + (m2)(g)(30) = 0.5(m1)(v1)^2 + 0.5(m2)(v2)^2 + (m2)(g)(16) Can you check if my eqn is correct? And can you advise what to do after this? I wanted to do COLM but i don't know what is the initial part.

The displacement of the particle is; s= ∫v dt s= ∫4-3t^2 dt s=4t-t^3+c When the particle is at the Origin, t=0; 0=4(0)-(0)^3+c c=0 So this becomes; s=4t-t^3 The particle next passes through the origin when; 4t-t^3=0 Factor out the common term -t; -t(t^2-4)=0 Rewrite t^2-4 as t^2-2^2 and factor...

I need to find a trajectory x = 4cos(2t) + 3sin(2t) y = 3cos(2t) - 4sin(2t)

Genevieve the galah wants to test out her new creation - the Nifty Numbat Launcher. It consists of a large, ideal spring with equilibrium length l1 = 1.7 m and spring constant k=41 N/m. The spring rests upon a ramp of length l1 inclined at an angle β=30∘ to the horizontal. The spring is attached...

Would the A = 2.0 ms^(-5/2) affect my answers for parts (a) to (c)? Since I basically ignored the unit ms^(-5/2) and integrated the function to get the ones for velocity and displacement. And as for part (d) explain the dimensions of A; I assumed that the ms^(1/2) comes from the square root due...

I am confused because the question implies that I need to do some sort of calculation with Kepler's law. I got ##r+d = \sqrt[3]{\frac{T^2 GM}{4 \pi^2} } ## But don't understand why I need this, since I already have the distance and the angular diameter should be ##\arctan (2R/d)## I think I...

I solved it using parallelogram law if vector addition but didn't got the correct answer.why? Is their any other way to add velocity vectors. How to do this problem

Hi, I have just joined the forum. Thank you all for being a part of such places so that people like me can get answers to the questions on their minds! --------------------------- I have been trying to understand how a quadcopter yaws. Referring to the figure below which is bird's eye view of...

Hi folks, See below for a solved question finding the down slope distance of an arrow. How easy would it be to solve this question by making the x-axis the slope direction?

I'm not sure about my answer

for example take a look on this graph

Is there any geometric approach??

stuck on this question

x

I tried doing (vi)(∆ t) / 1/2(a)(∆ t) = 1/2(a)( ∆ t)² / 1/2(a)(∆ t) That allows you to cross out the two accelerations out and you end with an answer of 20. But apparently, they never meet. I then tried making them into two different equations and using graphing technology (Desmos) and they...

Hi everyone I have an issue with my physics teacher about a question on displacement. I have inserted the screenshot of the question below. By the definition of displacement, I was sure, and still am sure my answer is correct. But my physics teacher insisted that I am wrong. Please, can anyone...

Here is my attempt at the solution but I got the wrong answer. The right answer is t=16.6s. I know from the book (this is an example problem) that the motorcycle reaches its max speed at t=7.0s. But I don’t know where I made the mistake that is causing me to get the wrong answer afterwords.

Question is extracted from "Ellad B Tadmor, Ronald E Miller, Ryan S Elliott - Continuum mechanics and thermodynamics From fundamental concepts to governing equations". I just got stuck at part (a). I think if part(a) is solved, I may be able to do the other parts.

5 links (thus 6 frames) and corresponding theta, alpha, d and a values are all given. However, I'm not sure how to start coding the DH parameters to derive a forward kinematics model. I know that I'll have to use matrix operators, but do I just put in the values of the 4 factors mentioned above...

We know that v=v(x, t) i.e. V is a function of time as well as space coordinate here I have only taken 1D motion for simplicity. By chain rule a=dv/dt= Dv/Dx*dx/dt +Dv/Dt*dt/dt where D /Dx represent partial differentiation along x coordinate. So, a=Dv/Dx*dx/dt +Dv/Dt Here we say that Dv/Dt=0...

I initially thought it is v, because the speed shouldn't change when the tape is on top.