(a) Displacement = 6.25
(b) Total distance traveled = 14.25
For the displacement, you need to calculate the integral for the given function to create the position function. Use the reverse power rule to do that.
v(t) = t^3 - 9t^2 + 23t - 15
P(t) = (1/4)t^4 - 3t^3 + 11.5t^2 - 15t + C
Now calculate P(6) - P(1)
P(6) = (1/4)6^4 - 3*6^3 + 11.5*6^2 - 15*6 + C
P(6) = (1/4)1296 - 3*216 + 11.5*36 - 15*6 + C
P(6) = 324 - 648 + 414 - 90 + C
P(6) = 0 + C
P(1) = (1/4)1^4 - 3*1^3 + 11.5*1^2 - 15*1 + C
P(1) = (1/4)1 - 3*1 + 11.5*1 - 15 + C
P(1) = (1/4) - 3 + 11.5 - 15 + C
P(1) = -6.25 + C
P(6) - P(1) = (0 + C) - (-6.25 + C) = 0 + C + 6.25 - C = 6.25
So the displacement is 6.25
(b) For the total distance traveled, we that the particle doesn't just simply move from -6.25 straight to 0, it moves back and forth a bit. So you need to find those points where the velocity changes sign. This can only happen at those points where the velocity is 0. So let's find the roots of v(t) = t^3 - 9t^2 + 23t - 15.
t = 1 is obviously a root, then doing the division gives t^2 - 8t + 15 which in turn factors into (t-3) and (t-5). So the roots are t=1, t=3, and t=5.
So let's add up the displacements for t=1 to 3, t=3 to 5, and t = 5 to 6. We've already calculated P(1) and P(6) above, so we just need P(3) and P(5)
P(3) = (1/4)3^4 - 3*3^3 + 11.5*3^2 - 15*3 + C
P(3) = (1/4)81 - 3*27 + 11.5*9 - 15*3 + C
P(3) = 20.25 - 81 + 103.5 - 45 + C
P(3) = -2.25 + C
P(5) = (1/4)5^4 - 3*5^3 + 11.5*5^2 - 15*5 + C
P(5) = (1/4)625 - 3*125 + 11.5*25 - 15*5 + C
P(5) = 156.25 - 375 + 287.5 - 75 + C
P(5) = -6.25 + C
d = abs(P(3) - P(1)) + abs(P(5) - P(3)) + abs(P(6)-P(5))
d = abs((-2.25 + C) - (-6.25 + C)) + abs((-6.25 + C) - (-2.25 + C)) + abs((0 +
C)-(-6.25 + C))
d = abs(-2.25 + C + 6.25 - C) + abs(-6.25 + C + 2.25 - C) + abs(0 + C + 6.25 - C)
d = abs(4) + abs(-4) + abs(6.25)
d = 4 + 4 + 6.25
d = 14.25
So the particle traveled a total distance of 14.25 units.
