The velocity of a particle moving in the x-y plane is given by (6.07i+ 4.97j) m/s at time t = 3.87 s. Its average acceleration during the next 0.014 s is (7.5i + 5.0j) m/s². Determine the velocity v of the particle at t = 3.884 s and the angle between the average- acceleration vector and the velocity vector at t = 3.884 s.

The velocity of a particle moving in the x-y plane is given by (6.07i+ 4.97j) m/s at time t = 3.87 s. Its average acceleration during the next 0.014 s is (7.5i + 5.0j) m/s². Determine the velocity v of the particle at t = 3.884 s and the angle between the average- acceleration vector and the velocity vector at t = 3.884 s.

Name: The velocity of a particle moving in the x-y plane is given by (6.07i
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: The velocity of a particle moving in the x-y plane is given by (6.07i+ 4.97j) m/s at time t = 3.87 s. Its average acceleration during thenext 0.014 s is (7.5i + 5.0j) m/s². Determine the velocity v of the particle at t = 3.884 s and the angle betwee

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter1: Introduction And Vectors

Section: Chapter Questions

Problem 4OQ: What is the y component of the vector (3i8k) m/s? (a) 3 m/s (b) 8 m/s (c) 0 (d) 8 m/s (e) none of...

See similar textbooks

Similar questions

The velocity of a particle moving in thex-y plane is given by (4.03i + 7.36j) m/s at time t = 3.64 s. Its average acceleration during the next 0.020 s is (2.2i + 2.6j) m/s?. Determine the velocity v of the particle at t = 3.660 s and the angle e between the average- acceleration vector and the velocity vector at t = 3.660 s. Answers: v= ( i i+ i j) m/s e =
The equation r(t) = ( sin t)i + ( cos t)j + (t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. π Then write the particle's velocity at t= as a product of its speed and direction. The velocity vector is v(t) = (i+j+ k.
The velocity of a particle moving in the x-y plane is given by (7.481 +6.64)) m/s at time t = 6.53 s. Its average acceleration during the next 0.012 s is (5.8i + 4.3)) m/s². Determine the velocity v of the particle at t = 6.542 s and the angle between the average- acceleration vector and the velocity vector at t = 6.542 s. Answers: v=i i i+ j) m/s
The position r of a particle moving in an xy plane is given by ř seconds. In unit-vector notation, calculate (a) 7, (b) V , and (c) a for t = 3.00 s. (d) What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 3.00 s? Give your answer in the range of (-180°; 180°). (4.00r3 – 1.00t)î + (5.00 – 1.00r4)j with 7 in meters and t in (a) Number i i Units (b) Number ît i Units i (c) Number i i Units (d) Number i Units
The acceleration of a particle moving only on a horizontal xy plane is given by á = 3t i+4t j, where á is in meters per second-squared and t is in seconds. At t = 0, the position vector is 7 = (20 m) i+(40 m) j and the velocity vector v = (5 m/s) i+(2 m/s) j. Find expressions for the (a) x and (b) y component (in meters) of the position of the particle at time t. (c) What is the angle between its velocity and the positive direction of the x axis at time t? (а) х %3D 5t+20) i m Edit (b) y = Edit (c) e = Edit Click if you would like to Show Work for this question: Open Show Work
The velocity vector of a particle is given by: V = Vje-)[sin(wt)i + cos(wt)j] where, Vo = 12.1 m/s; T = 2.5 s; w = 5.2 rad/s Calculate the magnitude of the acceleration (in m/s2) at t = 7.4 s
The velocity of a particle moving in the x-y plane is given by (7.15i +4.12j) m/s at time t = 5.37 s. Its average acceleration during the next 0.022 s is (3.9i+ 6.9j) m/s2. Determine the velocity v of the particle at t = 5.392 s and the angle between the average- acceleration vector and the velocity vector at t = 5.392 s. Answers: v = ( i 0= i it i j) m/s
At time t = 0, the position vector of a particle moving in the x-y plane is r = 4.62i m. By time t = 0.011 s, its position vector has become (4.84i +0.54j) m. Determine the magnitude Vay of its average velocity during this interval and the angle 0 made by the average velocity with the positive x-axis. Answers: Vav 0= i i m/s
The acceleration of a particle moving only on a horizontal xy plane is given by d = 6tî + 7tj, where d is in meters per second-squared and tis in seconds. At t= 0, the position vector 7 = (15.0m)î + (35.0m)ĵ locates the paticle, which then has the velocity vector V = (6.40m/s)î + (3.30m/s)ĵ. At t = 5.60 s, what are (a) its position vector in unit-vector notation and %3D (b) the angle between its direction of travel and the positive direction of the x axis? (a) Number i j Units i (b) Number i Units
The velocity of a particle moving in the x-y plane is given by (6.98i + 6.68j) m/s at time t = 7.60 s. Its average acceleration during the next 0.014 s is (4.1i + 5.4j) m/s2. Determine the velocity v of the particle at t = 7.614 s and the angle θ between the average-acceleration vector and the velocity vector at t = 7.614 s.Answers:v = ( _____ i + _____ j) m/sθ = _____ °
The velocity of a particle moving in the x-y plane is given by (6.12i + 3.24j) m/s at time t = 3.65 s. Its average acceleration during the next 0.02 s is (4.0i + 6.0j) m/s². Determine the velocity v of the particle at t = 3.67 s and the angle between the average-acceleration vector and the velocity vector at t = 3.67 s. Answers: V = ( i 8= i it i j) m/s
the coordinates of an object moving in an xy plane vary with the time according to the eqations x=-7.85 sin w(omega)t, and y=4-7.85 cosw(omega)t. W(omega) is constant , x,y - in meters. And t in seconds. Write expressions for the possition,velocity and acceleration vectors of the object at any time t>0.