The position of a particle moving along an x axis isgiven by x = 14.0 t 2 -4.00

Question

The position of a particle moving along an x axis isgiven by x = 14.0t2 -4.00t3, where x is in meters andt is in seconds. Determine (a) theposition, (b) the velocity, and(c) the acceleration of the particle at t= 4.00 s. (d) What is the maximum positivecoordinate reached by the particle and (e) at whattime is it reached? (f) What is the maximumpositive velocity reached by the particle and (g)at what time is it reached? (h) What is theacceleration of the particle at the instant the particle is notmoving (other than at t = 0)? (i)Determine the average velocity of the particle between t =0 and t = 4.00 s.

Explanation / Answer

x=14t^2-4t^3 so v=28t-12t^2 and a=28-24t *at t=4 s. x=-32m v=-80m/s a=-68m/s^2 *It reach the maximum positive position when it velocity becomezero. v=28t-12t^2=0 so t=2,33s. so x=25,4m *It reach the maximum positive velocity when its accelerationbecome zero. so a=28-24t=0 so t=1,167s so v=16,33m/s *When it is not moving the velocity is zero. so t=2,33s a=-27,9m/s^2 *it is v=-32/4=-8m/s

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