Suppose that a particle's position is described by r(t) = ti+ (t + 1) k. Find the velocity vector for the particle at t = 1: v(1) = Find the acceleration vector for the particle at t = 1: a(t) (Answer in terms of t.)
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- A particle initially located at the origin has an acceleration of a=3.00jm/s2 and an initial velocity of vi=5.00im/s. Find (a) the vector position of the particle at any time t, (b) the velocity of the particle at any time t, (c) the coordinates of the particle at t = 2.00 s, and (d) the speed of the particle at t = 2.00 s.A particle is moving on xy plane, with x=15sin(2t), and y = 20cos(2t) Find the magnitude of the velocity as a function of timeSolving equations of motion Given an acceleration vector,initial velocity ⟨u0, v0, w0⟩, and initial position ⟨x0, y0, z0⟩, find thevelocity and position vectors, for t ≥ 0. a(t) = ⟨sin t, cos t, 1⟩, ⟨u0, v0, w0⟩ = ⟨0, 2, 0⟩,⟨x0, y0, z0⟩ = ⟨0, 0, 0⟩
- A particle starts from the origin at t=0, with an initial velocity having an x component of 20m/s and a y component of -15m/s. The particle moves in the xy plane with an x component of acceleration only, given by Ax =4.0 m/s^2 Determine the components of the velocity vector at anytime and the total velocity vector at any time . Calculate the velocity and speed of the particle at t=5.0s.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).The vector position of a particle varies in time according to the expression r (3.00i 6.00t 2 j) m. (a) Findexpressions for the velocity and acceleration as functions of time. (b) Determine the particle’s position andvelocity at t 1.00 s
- The positionr→of a particle moving in an xy plane is given byr→=(5.00t3−6.00t)i^+(2.00−2.00t4)j^withr→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).From the origin, a particle starts at t = 0 s with a velocity vecv=7.0hatim/s and moves in the xy plane with a constant acceleration of a=(-9.0hati+3.0hatj)m/s^(2). At the time the particle reaches the maximum x coordinate, what is its (a) velocity and (b) position vector?if the marble is launched with the launch speed V_{o} but at an initial angle of , what would be the expression of v_{0} in terms of horizontal displacement x and vertical displacement y? Hint: y = (v_{0} * sin theta) * t - 1/2 * q * t ^ 2 and replace t using x = (v_{0} * cos theta) t .
- Let the position vector (with its tail at the origin) of a moving particle be~r(t) = t^(2)ˆi − 2tˆj + (t^(2) + 2t)ˆk, where ~r(t) is measured in meters and t is measured in seconds.(a) Find the acceleration vector and the magnitude of the particle at time t.(b) Find the acceleration vector and the magnitude of the particle when it passes throughthe point (4, -4, 8).A child running in a field has the following position and velocity at time t=0: x = 1.5 m. v= -2.0 m. Vx =-3.5 m/s, vy = 1.0 m/s At time t = 2.5 s, it has the following position and velocitv: x = -1.0 m, y = 1.5 m 08. Vx = -0.50 m/s, Vy = 2.0 m/s Using this information, find the magnitude and direction of the child's…A particle initially located at the origin has an acceleration of = 2.00ĵ m/s2 and an initial velocity of i = 9.00î m/s. (a) Find the vector position of the particle at any time t (where t is measured in seconds).( t î + t2 ĵ) m(b) Find the velocity of the particle at any time t.( î + t ĵ) m/s(c) Find the coordinates of the particle at t = 3.00 s.x = my = m(d) Find the speed of the particle at t = 3.00 s. m/s