A particle moves along the x-axis so that its velocity at time t is given by v(t)= (t^6-13t^4+12)/(10t^3+3). At time t=0, the initial position of the particle is x=7.

Question

(a) Find the acceleration of the particle at time t = 5.1

(b) Find all values of t in the interval 0 ≤ t ≤ 2 for which the speed of the particle is 1

(c) Find the position of the particle at time t = 4. Is the particle moving toward the origin or away from the

origin at time t = 4 ? Justify your answer.

(d) During the time interval 0 < t ≤ 4, does the particle return to its initial position? Give a reason for your answer

Answer 1

To find the acceleration function, take the derivative of the velocity function v(t). Acceleration is the rate of change of the velocity with respect to time.

Solve v(t) = 1 to find any values of t in the interval [0, 2] for which this is true.

To find the position function, you can integrate the velocity function, since velocity is the rate of change of position with respect to time.

The particle's initial position occurs at t = 0. Once you obtain the position function, evaluate it at t = 0, and see if there is any other value of t over the interval (0, 4] for which the position function has the same value that it does at t = 0.