Answer: a) d(t) = [tex]\frac{13}{14}[/tex] (t - 70) + 65 b: 6:41 pm
duration of time passed:
5:10 pm - 4:00 pm = 70 mins
distance = 65 miles
you points to use to make an equation:
(0,0) and (70,65) ---------------------> ( (0,0) is bc youre beginning the trip and you have no distance )
find the slope of your "future" equation
[tex]\frac{y_2-y_1}{x_2-x_1}[/tex] = [tex]\frac{65}{70}[/tex] = [tex]\frac{13}{14}[/tex] = m
now use point-slope form
y - 65 = [tex]\frac{13}{14}[/tex] (x - 70)
y= [tex]\frac{13}{14}[/tex] (x - 70) + 65
and since it wants it in terms of d and t,
d(t) = [tex]\frac{13}{14}[/tex] (t - 70) + 65
part b:
150 = [tex]\frac{13}{14}[/tex] (t - 70) + 65
85 = [tex]\frac{13}{14}[/tex] (t - 70)
85 = [tex]\frac{13}{14}[/tex] t - 65
150 = [tex]\frac{13}{14}[/tex] t
2100 = 13t
161.54 = t
about 162 minutes
Thennnn:
162/60 = 2 hours and 41 mins
4:00 pm + 2 hours and 41 mins = 6:41 pm
