The position of a particle moving along an x axis is given by x = 15.0t2 - 6.00t

Question

The position of a particle moving along an x axis is given by x = 15.0t2 - 6.00t3, 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 = 6.00 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 = 6.00 s.

Explanation / Answer

We begin by writing the position, velocity and acceleration. x(t) = 15 t2 - 6 t3 v(t) =dx/dt = 30t - 18t2 a(t) =dv/dt = 30 - 36 t We can now evaluate for (a), (b), and (c)... x(6) = - 756 m v(6) = - 468 m/s a(6) = - 186 m/s2 The maximum possible position occurs when the velocity reaches zero. We can find when this occurs most easily and then find the positions at these times. v(t) = 0 = 30t - 18t^2 0 = (30 - 18 t) t t = 0, 5/3 x(0) = 0 x(5/3) = 13.88 m We can see that the answer we are interested in is the x(5/3) = 13.88 m The maximum possible velocity occurs when the acceleration is zero. As in the previous case, it’s easiest to first find when this occurs. a(t) = 0 = 30 - 36 t t = 5/6 s v(5/6 ) = 12.5 m/sec We can compute the acceleration when the velocity is zero at 5/3 sec a (5/3) = -60 m/sec^2 Finally, we can compute the average velocity. vavg = (x(6) - x(0))/ (6s - 0s) = -126m/s

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