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 made by the average velocity with the positive x-axis.
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- The acceleration of a particle is a = (5.00 i - 7.00 j) m/s^2. At t = 0, the particle leavesthe origin with an initial velocity v0 = (4.00 i + 1.00 j) m/s. (a) What is the y-coordinate of the particle when its x-coordinate reaches 2.00 m? (b) What are the magnitude and direction of the particle's velocity when its x-coordinate reaches 2.00 m? Draw this velocity vector.A train at a constant 61.0 km/h moves east for 25.0 min, then in a direction 54.0\deg east of due north for 24.0 min, and then west for 35.0 min. What are the (a) magnitude and (b) angle (relative to east) of its average velocity during this trip?Suppose the position vector for a particle is given as a function of time by r(t) = x(t)î + y(t)ĵ, with x(t) : c = 0.120 m/s², and d = 1.02 m. (a) Calculate the average velocity during the time interval from t = 2.10 s to t = 3.85 s. V = m/s (b) Determine the velocity at t = 2.10 s. m/s = at + b and y(t) = ct² + d, where a = 1.80 m/s, b = Determine the speed at t = 2.10 s. m/s 1.20 m,
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