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# Question 2

last edited by 16 years, 5 months ago

# 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 + Co

• We can find Co 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 +Co
v(1) = 6(1)2 - 18(1) + Co
0 = 6 - 18 + Co
Co = 0 + 18 - 6
C= 6

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

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),

• Use the quadratic formula to find the roots.

x = -b + √(b2 - 4ac)
2a

= 3 + √((-3)2 - 4(1)(1))

2(1)
= 3 + √(5)
2

+                   -                         +

• v(t)                         +                               +

3-√(5)               3+√(5)

2                   2

•  And here we can see it graphically (c) Answer x(t) = 2t3 - 9t2 + 6t + 10

• Like part (a) we can find the equation of the x(t) by finding the antiderivative of v(t), or differentiating a(t) twice.

x(t) = S (6t2 - 18t +6)

= 2t3 - 9t2 + 6t +C1

• We can find C1 given the information x(1) = 9.

x(t) = 2t3 - 9t2 + 6t + C1

x(1) = 2(1)3 - 9(1)2 + 6(1) + C1
9 = 2 - 9 + 6 + C1
C1 = 10

• Therefore, the expression for x(t) of the particle is  x(t) = 2t3 - 9t2 + 6t + 10

• 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

= [648 - 648 + 108 + 60] - [2.53125 - 10.125 + 6.75 + 15]

= 168 - 14.15625

= 153.8438 units

• 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 