Position, Velocity, & Acceleration

Suppose a particle is moving on the x-axis. Its position x(t) at time t is x(t) = 2t3 - 13t2 + 22t - 5. Time is measured in seconds, and the particle will investigate the interval 0 < t < 5.

  1. When is the acceleration of the particle 0?
    1. v(t) is the derivative of x(t)
    2. a(t) is the derivative of v(t)
    3. v(t) = 6t2 - 26t + 22
    4. a(t) = 12t - 26
    5. Solve 12t - 26 = 0
                           t = 13/6 sec
  2. When is the velocity 0?
    1. v(t) = 6t2 - 26t + 22
    2. 6t2 - 26t + 22 = 0
      3t2 - 13t + 11 = 0
      Use your calculator or the quadratic formula to find the zeros.
      t = 3.180 sec and t = 1.153 sec
  3. When is the particle moving to the right?
    1. It moves to the right when v(t) > 0 and to the left when v(t) < 0.  It is not moving when v(t) = 0.
    2. v(t) = 6t2 - 26t + 22
    3. 6t2 - 26t + 22 > 0
      Solve, using your calculator or the quadratic formula (along with the squiggly line)
      t < 1.153 sec and t > 3.180 sec
  4. When is the particle slowing down?
    1. It slows down when the velocity and the acceleration are opposite in signs.
    2. v(t) is negative when 1.153 < t < 3.180
    3. a(t) = 12t - 26 so a(t) > 0 when t > 13/6 and a(t) < 0 when t < 13/6
    4. They are opposite in signs when 1.153 < t < 13/6, so the particle is slowing down in that time interval.
  5. When is the particle farthest from the origin?
    1. The only positions that need to be checked are the beginning, the turning points, and the endpoint.
    2. Beginning:  x(0) = -5
    3. End:  x(5) = 30
    4. Turning points occur when v(t) = 0 and changes sign.
      v(t) = 0 when t = 1.153 and t = 3.180
      x(1.153) = 6.149
      x(3.180) = -2,186
    5. 30 is farther from the origin than the other three positions, so the particle is farthest from the origin at t = 5 sec.
  6. When does the particle reach maximum speed?
    1. Maximum speed occurs when |v(t)| is the greatest.
    2. Maximum speed occurs either at the endpoints or when a(t) = 0 and changes sign.
    3. Endpoints:  |v(0)| = 22
                        |v(5)| = 42
    4. a(t) = 0 when t = 13/6, so |v(13/6)| = 6.167
  7. Find the distance traveled by the particle.
    1. Endpoints and turning points are again the important points.  The particle may change direction when v(t) = 0 and thus retrace old ground.  This retracing must be added into total distance covered.  
    2. x(0) = -5
    3. x(1.153) = 6.149
    4. x(3.180) = -2,186
    5. x(5) = 30
    6. Put them in order of t-values and subtract to find the distance traveled over each interval.
      x(1.153) - x(0) = 11.149
      x(3.180) - x(1.153) = 8.335
      x(5) - x(3.180) = 32.186
    7. Total distance is 11.149 + 8.335 + 32.186 = 51.67 units on the x-axis.