(a) If the position of a particle is given by x + 20t - 5t3, where x is in meters and t is in seconds, when, if ever, is the particle’s velocity zero? (b) When is its acceleration a zero? (c) For what time range (positive or negative) is a negative? (d) Positive? (e) Graph x(t), v(t), and a(t).

The velocity (in m/s) of a particle is given by v(t) = 2t- 6 where t is the time in seconds. During the interval 0

The position of a particle moving along an x axis is given by x = 13.0t2 - 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 = 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.(a).=-176m(b).=-184m/s (c).=-118.0m/s^2 (d).=9.041m (e).=1.444s (f).=9.389m/s (g).=0.722sPlease help me with h and i.

In physics, one important application of the derivative involves the height of a free – falling object, which can be modeled using the position function. where h is the height of the object in feet, t is the time in seconds, vo is the initial velocity (in feet per second), and ho is the initial height.

The velocity of a particle moving along the x-axis varies in time according to the expression v(t) = α - βt2 where α = 52.9m/s , β = 3.72m/s3,  and t is in seconds.  a) Find the acceleration in the time interval from t = 0 to 2.97s in units of m/s2 b) Determine the acceleration of the particle tf = 2.97s in m/s2