At t = 0, a particle moving with constant acceleration in the xy plane has a velocity v = (3.00i-2.00j) m / s at its origin. At t = 3.00 s, the velocity of the particle is v = (9.00i + 7.00j) m / s. Find the acceleration of the particle?

At t = 0, a particle moving with constant acceleration in the xy plane has a velocity v = (3.00i-2.00j) m / s at its origin. At t = 3.00 s, the velocity of the particle is v = (9.00i + 7.00j) m / s. Find the acceleration of the particle?

Name: 22)))At t = 0, a particle moving with constant acceleration in the xy
Uploaded: 2024-03-20T06:57:47.000Z
Channel: Bartleby
Description: 22)))At t = 0, a particle moving with constant acceleration in the xy plane has a velocity v = (3.00i-2.00j) m / s at its origin. At t = 3.00 s, the velocity of the particle is v = (9.00i + 7.00j) m / s. Find the acceleration of the particle?

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

Chapter3: Motion In Two Dimensions

Section: Chapter Questions

Problem 6P: At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of...

See similar textbooks

Similar questions

At t = 0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xyplane with a constant acceleration of (2.0i −4.0j) m/s2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle (in m/s)?
The position of a particle moving in the xy plane varies with time, and its coordinates are given by the following expressions: x(t) = 4.00 m +r cos[(4.00/s)t] and y(t) = r sin[(4.00/s)t], where r = 2.00 m, and x and y will be in meters when t is in seconds. Determine the following for this particle. (a) speed of the particle at any time m/s (b) magnitude of the acceleration of the particle at any time m/s?
The velocity i of a particle moving in the xy plane is given by 7 = ( 5.50 - 5.00 ji + 9.00 3. with V in meters per second and t (> 0) in seconds. At I- 1.40 s and in unit-vector notation, what are (a) the x component and (b) the y component of the Eceleratiun? (c) Whien(if uver) is the acceluration zero? (d) At wlat pusitive time does the speed equal 10.0 mh? (a) Numher 7 Unit (b) Number 1 Unit {c) Number Unit (d) Number Unit
0, y 0) 44. A projectile is launched from the point (x QC with velocity (12.0î + 49.0 j) m/s, at t = 0. (a) Make a table listing the projectile's distance |r| from the ori- gin at the end of each second thereafter, for 0 s I S 10 s. Tabulating the x and y coordinates and the compo- nents of velocity v and v will also be useful. (b) Notice that the projectile's distance from its starting point increases with time, goes through a maximum, and starts to decrease. Prove that the distance is a maximum when the position vector is perpendicular to the veloc- ity. Suggestion: Argue that if v is not perpendicular to f, then r|must be increasing or decreasing. (c) Determine the magnitude of the maximum displacement. (d) Explain
A particle moves through 3-space in such a way that its velocity is v(t) = i + tj + t ²k Find the coordinates of the particle at time t = 1 given that the particle is at the point (−1, 2, 4) at time t = 0.
A roller coaster is moving at a certain path in the x-y plane and can be considered as a particle with the velocity given by 6.12i+3.24j m/s at time t-3.65 sec. Its average acceleration during the next 0.02 seconds is 4i+6j m/s?. Determine the velocity v of the particle at t-3.67sec Ov=-6.201-3.36j m/s Ov-6.20i+3.36j m/s Ov=6.20i-3.36j m/s Ov=-3.25i-4.16j m/s Ov-3.25i+4.16j m/s Ov=-3.25i+4.16j m/s
(4.88î) á = (– 1.53î – 3.93j) m/s 2.when the particle reaches its maximum x coordinate, what are (a) its velocity, (b) its position A particle leaves the origin with an initial velocity v m/s and a constant acceleration vector? (a) Number ! ) Units (b) Number i i Units >
The velocity of a particle is given by v = {16t’i + 4t°j + (5t + 2)k} m/s, where t is in seconds. If the particle if at the origin when t = 0, determine the z-coordinate position %3D of the particle when t = 2 s.
A particle moves through 3-space in such a way that its acceleration is a(t)= 8sin 2ri +8cos 2tj+e'k . where t is time measured in seconds. The initial velocity of the particle is v, = 2i – 3j+k. At time t = t , find i) the velocity vector of the particle. ii) the scalar tangential and normal components of acceleration.
A particle leaves the origin with its initial velocity given by v=21.3i +14j m/s, undergoing constant acceleration a=-1.2 i +0.26j m/s?. When does the particle cross the y axis? Answer: Choose... →
A particle moving in the x-y plane has a velocity v = 7.80i + 2.97j m/s at a certain instant. If the particle then encounters a constant acceleration a = 0.88j m/s², determine the amount of time which must pass before the direction of the tangent to the trajectory of the particle has been altered by 33º. Answer: t = i S
At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of v; = (3.00 i – 2.00 j) m/s and is at the origin. At t = 2.20 s, the particle's velocity is v = (6.60 i + 8.80 j) m/s. (Round your coefficients to two decimal places.) (a) Find the acceleration of the particle at any time t. (Use the following as necessary: t.) a = 1.64i + 4.91j m/s? (b) Find its coordinates at any time t. (Use the following as necessary: t.) X = y = m