Speed and Total Distance

Speeding up and slowing down

Example 1

A particle's position is

x(t) = \ln\left(t^2 - 4t + 8\right)

meters for

t \ge 0

seconds. (a) Find its speed at

t = 1

. (b) Find the intervals on which the particle is speeding up and the intervals on which it's slowing down.

The chain rule gives

v(t) = \frac{2t - 4}{t^2 - 4t + 8}

. In the quotient rule for

a(t)

, the numerator

2\left(t^2 - 4t + 8\right) - (2t - 4)^2

simplifies to

-2t^2 + 8t

, so

a(t) = \frac{-2t(t - 4)}{\left(t^2 - 4t + 8\right)^2}

(a)

v(1) = -\frac{2}{5}

, so the speed is

|v(1)| = \frac{2}{5} = 0.4

meter per second.

(b) Both denominators are positive, since

t^2 - 4t + 8 = (t - 2)^2 + 4

. The signs come from

t - 2

and

-2t(t - 4)

Interval	$v(t)$	$a(t)$	Speed
$0 < t < 2$	$-$	$+$	decreasing
$2 < t < 4$	$+$	$+$	increasing
$t > 4$	$+$	$-$	decreasing

The particle is speeding up on

2 < t < 4

and slowing down on

0 < t < 2

and

t > 4

The speed falls from 1/2 to 0 at t = 2, rises to 1/2 at t = 4, then falls again.

Example 2

The graph shows the velocity

v(t)

, in feet per second, of a particle for

0 \le t \le 6

seconds. It's made of line segments.

The velocity crosses zero at t = 1 and t = 4.

(a) Find the speed of the particle at

t = 1.5

and at

t = 4.5

. (b) Find the intervals on which the speed is increasing, decreasing, and constant.

(a) On

0 \le t \le 2

v(t) = 2 - 2t

, so

v(1.5) = -1

and the speed is

1

foot per second. On

3 \le t \le 5

v(t) = 2t - 8

, so

v(4.5) = 1

and the speed is again

1

foot per second.

(b) The acceleration is the slope of each segment:

-2

(0, 2)

0

(2, 3)

2

(3, 5)

, and

0

(5, 6)

\begin{aligned} &\text{increasing: } (1, 2) \text{ and } (4, 5), \\ &\text{decreasing: } (0, 1) \text{ and } (3, 4), \\ &\text{constant: } (2, 3) \text{ and } (5, 6). \end{aligned}

(1, 2)

v

and

a

are both negative, and on

(4, 5)

both are positive. On

(0, 1)

and

(3, 4)

they have opposite signs.

Total distance traveled

Example 3

A particle's position is

x(t) = 3t^4 - 16t^3 + 18t^2

feet for

0 \le t \le 4

seconds. (a) Find the displacement and the total distance traveled over

0 \le t \le 4

. (b) Find where the particle is farthest left and farthest right on that interval.

(a) The velocity

v(t) = 12t^3 - 48t^2 + 36t

factors as

12t(t - 1)(t - 3)

. Inside the interval it changes sign at

t = 1

and

t = 3

. The positions there and at the ends are

\begin{aligned} x(0) &= 0, & x(1) &= 5, \\ x(3) &= -27, & x(4) &= 32. \end{aligned}

The displacement is

x(4) - x(0) = 32

feet. The total distance is

\begin{aligned} &|5 - 0| + |-27 - 5| + |32 - (-27)| \\ &\qquad = 5 + 32 + 59 = 96 \text{ feet}. \end{aligned}

The path along the x-axis: right from 0 to 5, left to −27, then right to 32. The legs are 5, 32, and 59 feet.

(b) Between turns the particle travels in one direction, so the extreme positions occur at a turn or an end. Of

0

5

-27

, and

32

, the particle is farthest left at

t = 3

, at

x = -27

, and farthest right at

t = 4

, at

x = 32

Example 4

A buoy bobs on the waves. Its height above its resting level is

h(t) = 4\cos\left(\frac{\pi t}{3}\right)

inches at time

t

seconds. Find the total distance the buoy travels over

0 \le t \le 7

The velocity is

h'(t) = -\frac{4\pi}{3}\sin\left(\frac{\pi t}{3}\right)

, which is

0

t = 0, 3, 6

. It's negative on

(0, 3)

, positive on

(3, 6)

, and negative on

(6, 7)

, so the buoy turns at

t = 3

and

t = 6

. With

h(0) = 4

h(3) = -4

h(6) = 4

, and

h(7) = 2

\begin{aligned} &|-4 - 4| + |4 - (-4)| + |2 - 4| \\ &\qquad = 8 + 8 + 2 = 18 \text{ inches}. \end{aligned}

The height h(t): down 8 inches, up 8 inches, then down 2 inches.