Answer
(a) Jari
(b) Jade: $60 \space mi/h$ | Jari: $70 \space mi/h$
(c) $f(t) = t + 10$ and $g(t) = \frac 7 6 t $
Work Step by Step
(a)
- From t = 0 to t = 6, Jade traveled 6 miles, and Jari traveled 7 miles. Therefore, $Jari$ is moving faster.
(b)
Jade: $speed = a = \frac{16 - 10}{6 - 0} = \frac 6 6 = \frac{1 \space mi}{min} \times \frac{60 \space min}{1 \space h} = 60 \space mi/h$
Jari : $speed = a = \frac{7 - 0}{6 - 0} = \frac {7 \space mi} {6 \space min} \times \frac{60 \space min}{1 \space h} = 70 \space mi/h$
(c)
** Notice: $a$ must be in $\frac{mi}{min}$
Jade: f(t) = at + b
At 7:00 A.M. Jade had already traveled 10 miles. b = 10
$$f(t) = (1)t + 10 = t + 10$$
Jari: g(t) = at + b
At 7:00 A.M. Jari started moving. b = 0
$$g(t) = \frac 7 6 t $$