Question 2

You MAY use a calculator to solve this problem.

A particle moves along the x-axis so that its acceleration at any time t > 0 is given by a(t) = 12t - 18. At time t = 1, the velocity of the particle is v(1) = 0 and the position is x(1) = 9.

(a) Write an expression for the velocity of the particle v(t).

(b) At what values of t does the particle change direction?

(c) Write an expression for the position x(t) of the particle.

(d) Find the total distance traveled by the particle from t = 3/2 to t = 6.

Solution

(a) Answer v(t) = 6t² - 18t + 6

• We can find the equation of the v(t) by finding the antiderivative of a(t).
   
  a(t) = 12t - 18

  v(t) = S (12t - 18) dx

  v(t) = 6t² - 18t + C_o

   
• We can find C_o because we are given the information v(1) = 0 so that pins it down to one answer.  Plug in the given values into the equation.
   

  v(t) = 6t² - 18t +C_o

  v(1) = 6(1)² - 18(1) + C_o

  0 = 6 - 18 + C_o

  C_o = 0 + 18 - 6

  C_o = 6

   
• Therefore, the expression for v(t) is v(t) = 6t² - 18t + 6

(b) Answer t = (3±√(5)) 

                               2

• The particle on a velocity-time graph chages direction ONLY IF, there is a sign change from the left to the right of a root.
   
• Wherever the velocity-time graph is negative, the particle is moving away from its reference point in a backwards direction.  If it's positive, the particle is moving forward.

• Since x'(t) = v(t), we can illustrate what's happening using a number line of v(t).
   
   
  v(t) = 6t² - 18t + 6
  v(t) = 6 (t² - 3t + 1),

                       +                   -                         +

• v(t)                         +                               +                          

                             3-√(5)               3+√(5)

                           2                   2

•  And here we can see it graphically

(d) Answer 153.8438 units.

• Integrating a position-time graph within an interval will give you the total distance traveled for that specific time frame.

                    6

          a = § x(t)

                    3/2

                    6

             = § (2t³ - 9t² + 6t + 10) dx 

                    3/2

                                                    6                          

              = [(1/2)t⁴ - 3t³ + 3t² + 10t]

                                                    3/2

• We can also use the calculator by finding the signed are nder the curve over that interval.  You can use the fnInt( command, or by graphing it and using the 2nd Trace command.
  -manny