ea of Limits
The position of an object moving along a line is given by the function s(t)=-172 +102t. Find the average velocity of the object over the following
intervals.
(a) [1, 8]
(c) [1,6]
(b) [1,7]
(d) [1, 1+h] where h>0 is any real number.
(a) The average velocity of the object over the interval [1, 8] is



Answer :

To find the average velocity of an object moving along a line given by the function [tex]\( s(t) = -172 + 102t \)[/tex] over various intervals, follow these steps:

### Interval [1, 8]
First, calculate the average velocity over the interval [1, 8].

Step 1: Identify the position function [tex]\( s(t) \)[/tex].

[tex]\[ s(t) = -172 + 102t \][/tex]

Step 2: Evaluate the position at the beginning and end of the time interval.
- For [tex]\( t = 1 \)[/tex]:

[tex]\[ s(1) = -172 + 102 \cdot 1 = -172 + 102 = -70 \][/tex]

- For [tex]\( t = 8 \)[/tex]:

[tex]\[ s(8) = -172 + 102 \cdot 8 = -172 + 816 = 644 \][/tex]

Step 3: Compute the change in position [tex]\(\Delta s\)[/tex].

[tex]\[ \Delta s = s(8) - s(1) = 644 - (-70) = 644 + 70 = 714 \][/tex]

Step 4: Compute the change in time [tex]\(\Delta t\)[/tex].

[tex]\[ \Delta t = 8 - 1 = 7 \][/tex]

Step 5: Calculate the average velocity [tex]\(\bar{v}\)[/tex].

[tex]\[ \bar{v} = \frac{\Delta s}{\Delta t} = \frac{714}{7} = 102 \][/tex]

Therefore, the average velocity of the object over the interval [tex]\([1, 8]\)[/tex] is [tex]\( 102 \, \text{units of distance per unit time} \)[/tex].

### Interval [1, 7]
Repeat the steps for the interval [1, 7]:

Step 1: Evaluate the position at [tex]\( t = 1 \)[/tex] and [tex]\( t = 7 \)[/tex].
- For [tex]\( t = 1 \)[/tex]:

[tex]\[ s(1) = -172 + 102 = -70 \][/tex]

- For [tex]\( t = 7 \)[/tex]:

[tex]\[ s(7) = -172 + 714 = 542 \][/tex]

Step 2: Compute [tex]\(\Delta s\)[/tex].

[tex]\[ \Delta s = s(7) - s(1) = 542 - (-70) = 542 + 70 = 612 \][/tex]

Step 3: Compute [tex]\(\Delta t\)[/tex].

[tex]\[ \Delta t = 7 - 1 = 6 \][/tex]

Step 4: Calculate the average velocity.

[tex]\[ \bar{v} = \frac{\Delta s}{\Delta t} = \frac{612}{6} = 102 \][/tex]

Therefore, the average velocity over the interval [tex]\([1, 7]\)[/tex] is [tex]\( 102 \, \text{units of distance per unit time} \)[/tex].

### Interval [1, 6]
Repeat the process for the interval [1, 6]:

Step 1: Evaluate the position at [tex]\( t = 1 \)[/tex] and [tex]\( t = 6 \)[/tex].
- For [tex]\( t = 1 \)[/tex]:

[tex]\[ s(1) = -70 \][/tex]

- For [tex]\( t = 6 \)[/tex]:

[tex]\[ s(6) = -172 + 612 = 440 \][/tex]

Step 2: Compute [tex]\(\Delta s\)[/tex].

[tex]\[ \Delta s = s(6) - s(1) = 440 - (-70) = 440 + 70 = 510 \][/tex]

Step 3: Compute [tex]\(\Delta t\)[/tex].

[tex]\[ \Delta t = 6 - 1 = 5 \][/tex]

Step 4: Calculate the average velocity.

[tex]\[ \bar{v} = \frac{\Delta s}{\Delta t} = \frac{510}{5} = 102 \][/tex]

Therefore, the average velocity over the interval [tex]\([1, 6]\)[/tex] is [tex]\( 102 \, \text{units of distance per unit time} \)[/tex].

### Interval [1, 1+h], where [tex]\( h > 0 \)[/tex]
For the general interval [tex]\([1, 1 + h]\)[/tex]:

Step 1: Evaluate the position at [tex]\( t = 1 \)[/tex] and [tex]\( t = 1 + h \)[/tex].
- For [tex]\( t = 1 \)[/tex]:

[tex]\[ s(1) = -70 \][/tex]

- For [tex]\( t = 1 + h \)[/tex]:

[tex]\[ s(1 + h) = -172 + 102(1 + h) = -172 + 102 + 102h = -70 + 102h \][/tex]

Step 2: Compute [tex]\(\Delta s\)[/tex].

[tex]\[ \Delta s = s(1 + h) - s(1) = (-70 + 102h) - (-70) = 102h \][/tex]

Step 3: Compute [tex]\(\Delta t\)[/tex].

[tex]\[ \Delta t = (1 + h) - 1 = h \][/tex]

Step 4: Calculate the average velocity.

[tex]\[ \bar{v} = \frac{\Delta s}{\Delta t} = \frac{102h}{h} = 102 \][/tex]

Therefore, the average velocity over the interval [tex]\([1, 1 + h]\)[/tex] is [tex]\( 102 \, \text{units of distance per unit time} \)[/tex].

In summary, for all given intervals, the average velocity of the object is [tex]\( 102 \, \text{units of distance per unit time} \)[/tex].