The position of a particle moving along an x axis is given by x = 14.0t2 - 4.00t

Question

The position of a particle moving along an x axis is given by x = 14.0t2 - 4.00t3, where x is in meters and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t = 4.00 s. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached? (f) What is the maximum positive velocity reached by the particle and (g) at what time is it reached? (h) What is the acceleration of the particle at the instant the particle is not moving (other than at t = 0)? (i) Determine the average velocity of the particle between t = 0 and t = 4.00 s.

Explanation / Answer

(a) x=14(4)2-4(4)3 =-32m (position cannot be negative) (b) Velocity=dx/dt=28t-12t2=28*4-12*42=-80 m/s2 (c) Acceleration=d2x/dt2=28-24t=28-24*4=-68 m/s2 (retedration) (d,e) Maximum positive coordinate x = 14.0t2 - 4.00t3 dx/dt=28t-12t2 for maximum and minimum point dx/dt=0 we get t=0 and t=28/12 d2x/dt2=28-24t at t=28/12 we get the negative value so we can say that at t=28/12 Sec the position is maximum=25.04 m (f) positive velocity=28t-12t2 at t=28/12 =.0932m/s (g) Acceleration =d2x/dt2=28-24t (f) Average velocity=(velocity at t=0+velocity at t=4)/2 =0-80/2=-40m/s2

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