Calculus Velocity: An Object moves with velocity v(t) = t^2 -8t +7?

Question

An Object moves with velocity v(t) = t^2 -8t +7

(A) write a polynomial expression for the position of the particle at any time t >= 0.

(B) At what time(s) is the particle changing direction?

(C) Find the total distance traveled by the particle from time t=0 to t=4.

Answer 1

a) the position is the integral of the velocity function,
say
s(t) = t^3/3 -4t^2 +7t + c

where c is a constant corresponding to the position at t=0.

b) the particle changes direction whenever the velocity is zero; the velocity function equals

(t-1)(t-7) a difference of squares so the zeros are 1 and 7, it changes direction at 1 second and 7 seconds.

c) s(0) = c; s(1) = 8/3 + c; s(4) = 64/3 - 64 + 28 + c.

from 0 to 1 the particle travels 8/3 units. From 1 to 4 it travels -(64/3 - 36 - 8/3) = (-(56/3 - 108/3))
=-(-52/3) = 52/3 units
so in total it travels 52/3 + 8/3 =20 units.

Answer 2

(A)
Position is found by integrating velocity:

s(t) = (t^3)/3 - 4t^2 + 7t + C



(B)
Particles change direction when v(t) = 0
v(t) = t^2 -8t +7 = 0
(t - 7)(t - 1) = 0
t = 1 and t = 7

(C)
Total distance is found by integrating the absolute value of the velocity function

Int {0 to 4} [ |t^2 - 8t + 7| ]

Use t = 1 and t = 7 to see on what intervals v(t) < 0
v(0) = 7, positive
v(2) = -5, negative
v(8) = 7, positive

v(t) = +(t^2 - 8t + 7) for 0 < t < 1
v(t) = -(t^2 - 8t + 7) for 1 < t < 4

So,
Int {0 to 4} [ |t^2 - 8t + 7| ] =

+Int {0 to 1} (t^2 - 8t + 7) + - Int {1 to 4} (t^2 - 8t + 7)
=[(1/3) - 4 + 7] - [(16 - 32 + 7) - ( (1/3) - 4 + 7)]
=(10/3) - [ -9 - (10/3) ]
=(10/3) + (37/3)
=47/3