Question

1. procedures

i. graphing and interpreting linear data
a. make a plot of the following data on a piece of graph paper, plotting position (x) vs time (t). if you do not have graphing paper you can download some samples. you should label the axes with proper units. draw a “best fit” line through the data by hand. the line should be as close to as many data points as possible. determine the slope (m) and y-intercept (b) of the line. write the equation of the line using slope intercept form, x(t) = mt + b.

t(s)	x(m)
1	21
2	29
3	41
4	49
5	60

m = __________, b = __________

write an expression for x(t) in slope-intercept form, x(t) = mt+b

x(t) = ____________

Explanation:

Step1: Find the y - intercept (b)

When \( t = 0 \), from the data table, \( x=10 \). In the equation \( x(t)=mt + b \), when \( t = 0 \), \( x(0)=m\times0 + b=b \). So \( b = 10 \).

Step2: Calculate the slope (m)

We can use two points to calculate the slope. Let's use the points \( (t_1,x_1)=(0,10) \) and \( (t_2,x_2)=(5,60) \). The formula for slope \( m=\frac{x_2 - x_1}{t_2 - t_1} \).
Substitute the values: \( m=\frac{60 - 10}{5 - 0}=\frac{50}{5}=10 \) (we can also check with other points, for example, using \( (0,10) \) and \( (1,21) \), \( m=\frac{21 - 10}{1 - 0}=11 \), but since we are drawing a best - fit line, we can take an average or a more representative slope. Let's recalculate with a better pair. Using \( (0,10) \) and \( (5,60) \) gives \( m = 10 \), using \( (1,21) \) and \( (4,49) \), \( m=\frac{49 - 21}{4 - 1}=\frac{28}{3}\approx9.33 \), but a more approximate best - fit slope: let's sum the differences. The change in \( t \) from 0 to 5 is 5, change in \( x \) is \( 60 - 10 = 50 \), so \( m = 10 \) is a good approximation.

Step3: Write the equation of the line

Using the slope - intercept form \( x(t)=mt + b \), with \( m = 10 \) and \( b = 10 \), the equation is \( x(t)=10t + 10 \). (Note: If we calculate the slope more accurately, for example, using the method of least squares:
The formula for the slope of the least - squares line for data points \( (t_i,x_i) \) is \( m=\frac{n\sum t_ix_i-\sum t_i\sum x_i}{n\sum t_i^2-(\sum t_i)^2} \) and \( b=\frac{\sum x_i - m\sum t_i}{n} \), where \( n = 6 \) (number of data points: \( t = 0,1,2,3,4,5 \); \( x = 10,21,29,41,49,60 \))
\(\sum t_i=0 + 1+2 + 3+4 + 5=\frac{5\times(5 + 1)}{2}=15\)
\(\sum x_i=10+21 + 29+41+49+60=210\)
\(\sum t_ix_i=0\times10+1\times21 + 2\times29+3\times41+4\times49+5\times60=0 + 21+58 + 123+196+300 = 698\)
\(\sum t_i^2=0^2+1^2+2^2+3^2+4^2+5^2=0 + 1+4 + 9+16+25 = 55\)
\(m=\frac{6\times698-15\times210}{6\times55 - 15^2}=\frac{4188-3150}{330 - 225}=\frac{1038}{105}\approx9.886\approx10\)
\(b=\frac{210-9.886\times15}{6}=\frac{210 - 148.29}{6}=\frac{61.71}{6}\approx10.285\approx10\))

Answer:

\( m=\boldsymbol{10} \), \( b=\boldsymbol{10} \)
\( x(t)=\boldsymbol{10t + 10} \)