The position of a particle moving along an x axis is given by x = 13t2 - 4.0t3,

Question

The position of a particle moving along an x axis is given by x = 13t2 - 4.0t3, where x is in meters and t is in seconds. (a) Determine the position, velocity, and acceleration of the particle at t = 3.0 s. x = m v = m/s a = m/s2 (b) What is the maximum positive coordinate reached by the particle? m At what time is it reached? s (c) What is the maximum positive velocity reached by the particle? m/s At what time is it reached? s (d) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? m/s2 (e) Determine the average velocity of the particle between t = 0 and t = 3 s. m/s

Explanation / Answer

x = 22t^2 - 3.0t^3 v = dx/dt = 44t - 9t^2 a = dv/dt = 44 - 18t a) t= 3 x = 117 m v = 51 m/s a = -10 m/s^2 b) dx/dt = 0 t = 44/9 x = 175.275 m c) maximum velocity at dv/dt =0 t = 44/18 v=53.778 m/s d) a = -73.33 m/s^2 e)avg velocity = x(3) - x(0) / 3 = 117/3 = 39 m/s

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