At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i − 2.00 j) m/s and is at the origin. At t = 2.60 s, the particle's velocity is v = (8.70 i + 3.40 j) m/s. (Round your coefficients to two decimal places.) Find its coordinates at any time t. (Use the following as necessary: t.)

At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi = (3.00 i − 2.00 j) m/s and is at the origin. At t = 2.60 s, the particle's velocity is v = (8.70 i + 3.40 j) m/s. (Round your coefficients to two decimal places.) Find its coordinates at any time t. (Use the following as necessary: t.)

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

A particle is moving on top of a 2-dimensional. plane with its coordinates given in cartesian system as x(t) =a sin ωt, y(t) = a cos ωt. Express the motion of the particle in terms of polar coordinates (ρ, φ). What is the minimum number ofgeneralised coordinates required to describe. its motion? Draw the. trajectory of the particle.
At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of i = (3.00 i − 2.00 j) m/s and is at the origin. At t = 2.30 s, the particle's velocity is = (7.60 i + 7.90 j) m/s. (Round your coefficients to two decimal places.) Find its coordinates at any time t. (Use the following as necessary: t.) Do not find coordinates at a specific point, but provide equations for finding any point of x and any point of y; ex. y=______ and x=_________.
The coordinates of a particle moving in the XY plane are given as function of time by: X = 5m + ( 7 m/s^4) t ^4 Y = ( 10 m/s)t + (3 m/s^5)t ^5 Find the position vector of the body at t = 3.5 seconds and instantaneous acceleration of the body at t = 3.5 sec
At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi=(3.00i2.00j)m/s and is at the origin. At t = 3.00 s, the particles velocity is vf=(9.00i+7.00j)m/s. Find (a) the acceleration of the particle and (b) its coordinates at any time t.
The position r→ of a particle moving in an xy plane is given by r→=(3.00t^3−7.00t)i^+(6.00−2.00t^4)j^ with r→ in meters and t in seconds. In unit-vector notation, calculate(a)r→, (b)v→, and (c)a→ for t = 2.00 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 = 2.00 s? Give your answer in the range of (-180o; 180o).
A particle is in uniform circular motion about the origin of an xy coordinate system, moving clockwise with a period of 8.40 s. At one instant, its position vector (from the origin) is r→=(3.70m)î-(4.50m)ĵ. At that instant, what is its velocity in unit-vector notation?
A particle which moves with curvilinear motion has coordinates in millimeters which vary with the time t in seconds according to x = 4.8t2 - 6.6t and y = 1.9t2 - t3/4.0. Determine the magnitudes of the velocity v and acceleration a and the angles which these vectors make with the x-axis when t = 1.3 s.Answers: When t = 1.3 s,
what is the resultant magnitude and directions? Given the vectors A= 6i + 9j, B = -9i + 9j, and C= 15i - 10j
What is the angle between the directions of vector A = 3.00i + 1.00j and vector B = -3.00i + 3.00j?
The acceleration of a particle moving only on a horizontal xy plane is given by a→=4ti^+5tj^, where a→ is in meters per second-squared and t is in seconds. At t = 0, the position vector r→=(28.0m)i^+(48.0m)j^ locates the paticle, which then has the velocity vector v→=(6.70m/s)i^+(2.30m/s)j^. At t = 4.60 s, what are (a) its position vector in unit-vector notation and (b) the angle between its direction of travel and the positive direction of the x axis?
A particle starts from the origin at t = 0 with a velocity of 6.0 m/s and moves in the xy plane with a constant acceleration of (−2.0î + 4.0ĵ ) m/s2 . At the instant the particle achieves its maximum positive x coordinate, how far is it from the origin?
If a spaceship traveling through deep space with the velocity vector as a function of time: v = (2t i +4t j+ 6t k) m/s. (time is in seconds). Given the initial position of the spaceship is r0= (20 i− 50 j − 100 k ) m. At what time would the spaceship be the closest to the origin? and What would the acceleration of the spaceship be at this instant? If you can express the acceleration in Cartesian vector form. Also ? is the unit vector in the z-direction.