Q1/ A partical has an acceleration gven by relation a = 0.3 V2v² + 36, where a and v are expressed in m/s² and m/s respectively, starting from x=0 with no initial velocity. 1) The position of the partical when v=4 m/s, is: our answer 1/b/the speed of the particle at X=3m is: * 3.524 2.543 2.98 1.654 4.21 4.85 Other: 1/c/the acceleration of the particle at X=3 mis: * 3.2399 2.3399

Q1/ A partical has an acceleration gven by relation a = 0.3 V2v² + 36, where a and v are expressed in m/s² and m/s respectively, starting from x=0 with no initial velocity. 1) The position of the partical when v=4 m/s, is: our answer 1/b/the speed of the particle at X=3m is: * 3.524 2.543 2.98 1.654 4.21 4.85 Other: 1/c/the acceleration of the particle at X=3 mis: * 3.2399 2.3399

Name: Your answerQ1/b/the speed of the particle at X=3m is: *оооQ1/ A parti
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: Your answerQ1/b/the speed of the particle at X=3m is: *оооQ1/ A partical has an acceleration gven by relation a = 0.3 √2v² + 36, where aand v are expressed in m/s² and m/s respectively, starting from x=0 with no initialvelocity.1) The position of th

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A particle moving along the x-axis has velocity function v(t) = t³ sin t. How far does the particle travel from time t = 0to t = π?
Here are a couple acceleration problems. Acceleration isn't necessarily constant. These problems will be pure math problems, meaning that we won't use units. a(t) = dv/dt = 146 cos(4t) v(t) = dx/dt, v(0) = -3, and x(0) = -6 Calculate x(6).
An object located at ?⃗0 = (1, 1) ? is displaced by Δ?⃗01 = −2?̂ − 5?̂ ? The object is then displaced by Δ?⃗12 = 3?̂ + 2?̂ ? a) Draw a picture b) What is the final position (?⃗2) of the object? c) How far did the object travel during each displacement? (?: 5.39 ?; 3.61 ?) d) What is the total distance travelled by the object? (?: 9 ?) e) How far did the object end up from where it initially started?(?: 3.16?)
A point particle of mass m = 1.8 kg moves according to the position function: r(t) = xtai + ytbj + ztck, where t denotes time and x, y, z, a, b, and c are constants such that the exponents are positive integers and the position function has the dimension of length. Part (a) We can write the particle’s velocity function in the form v(t) = ntdi + otej + ptgk. Enter an expression for n in terms of x, y, z, a, b, and c. Part (b) The particle’s velocity function will have the form v(t) = ntdi + otej + ptgk. Enter an expression for d in terms of x, y, z, a, b, and c. Part (c) Here is a set of parameter values for the motion of the particle: m = 1.8 kg, x = 1.8 m/s0, y = 2.4 m/s1, z = 0.15 m/s2, a = 0, b = 1, c = 2. Calculate the x-component of the particle’s angular momentum, in units of kg˙m2/s, about the origin at time t = 1 s. Part (d) Use the same set of parameter values (m = 1.8 kg, x = 1.8 m/s0, y = 2.4 m/s1, z = 0.15 m/s2, a = 0, b = 1, c = 2) to calculate the y-component of…
Two-dimensional motion: Object A has a position as a function of time given by rA(t) = (3.00 m/s)t i + (1.00 m/s2)t2j. Object B has a position as a function of time given by r p(t) = (4.00 m/s)ti + (-1.00 m/s2)t2i. All quantities are SI units. What is the distance between object A and object B at timet= 3.00 s?
The position-time graph that describes the motion of an object is given as in the figure below. if the values of a= 100.8m, b%3D212.8m and c=327.3m, Find the instantanous velocity (in units of m/s) at t=35 s. x(m) a t(s) 10 20 30 40 Select one: OA. 15.10 OB, 16.00 OC. 3.27 OD. 42.81 OE 22.69
٤٣ B/s ۱۰:۰۹ ص .in the direction of the force TO FO choose the correct answer The acceleration Vector in cartizan Cordinat writin by the formulla Ⓒazé x + j X+kŹ =¿Ÿ+K Ⓒă= cx+jŸ+k² @a=ix + ³y +kż a b с do time derivative of velocity is called acceleration rate velocity docs.google.com/forms/d/ तं
at an instant in time a particles velocity is v = (2.00i + 1.00j) m/s, while it’s acceleration is a = -1.00i m/s^2. at what rate is the particle speeding up / slowing down and what’s the radius of the curve on the particles trajectory
A particle moves in a curve path according to the law s = t^3 – 40t^2 + 8t + 100 where s is in meter and t is in sec. What is the normal component of acceleration of the particle if t = 5 seconds and the radius of the curve path is 10 meters?
A particle is moving in three dimensions and its position vector is given by; r→(t)=(1.6t^2+3.7t)i+(3.7t−1.4)j+(4.6t^3+3.8t)k where r is in meters and ?t is in seconds. Determine the magnitude of the average acceleration between t=0s and t=2s. Express your answer in units of m/s2 using one decimal place.
v(t) = 4t3 + 10t m/s. The initial position is r = 3 m. 3. A one dimensional problem. Assume v = Determine the position and acceleration at time t = 3sec.
PLEASE ANSWER QUICKLT AND WILL RATE HIGHLY *15