Question

1. the table gives the position ( s(t) ) of an object moving along a line at time ( t ), over a two-second interval. find the average velocity of the object over the following intervals.

( \boldsymbol{\text{a. } 0, 2} quad \boldsymbol{\text{b. } 0, 1.5} quad \boldsymbol{\text{c. } 0, 1} quad \boldsymbol{\text{d. } 0, 0.5} )

\begin{tabular}{|l|c|c|c|c|c|}
hline ( t ) & 0 & 0.5 & 1 & 1.5 & 2 \\
hline ( s(t) ) & 0 & 30 & 52 & 66 & 72 \\
hline
end{tabular}

Explanation:

Part a: Interval \([0, 2]\)

Step1: Recall average velocity formula

The formula for average velocity over an interval \([a, b]\) is \(\frac{s(b) - s(a)}{b - a}\). For the interval \([0, 2]\), \(a = 0\), \(b = 2\).

Step2: Substitute values from the table

From the table, \(s(0)=0\) and \(s(2) = 72\). Substitute into the formula: \(\frac{s(2)-s(0)}{2 - 0}=\frac{72 - 0}{2-0}\)

Step3: Calculate the result

\(\frac{72}{2}=36\)

Part b: Interval \([0, 1.5]\)

Step1: Apply average velocity formula

Using \(\frac{s(b)-s(a)}{b - a}\) with \(a = 0\), \(b = 1.5\).

Step2: Get values from table

\(s(0) = 0\), \(s(1.5)=66\). So, \(\frac{s(1.5)-s(0)}{1.5 - 0}=\frac{66-0}{1.5-0}\)

Step3: Compute the value

\(\frac{66}{1.5} = 44\)

Part c: Interval \([0, 1]\)

Step1: Use average velocity formula

For \([0, 1]\), \(a = 0\), \(b = 1\), formula is \(\frac{s(1)-s(0)}{1 - 0}\).

Step2: Substitute table values

\(s(0)=0\), \(s(1) = 52\). So, \(\frac{52-0}{1-0}\)

Step3: Calculate

\(\frac{52}{1}=52\)

Part d: Interval \([0, 0.5]\)

Answer:

s:
a. \(\boldsymbol{36}\)
b. \(\boldsymbol{44}\)
c. \(\boldsymbol{52}\)
d. \(\boldsymbol{60}\)