the function (s(t)=-\frac{60}{49}t^{2}+\frac{120}{7}t) models the speed (s) (in miles per hour) of a car (t)…

Explanation:

Step1: Identify the function type

The function $s(t)=-\frac{60}{49}t^{2}+\frac{120}{7}t$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a =-\frac{60}{49}$, $b=\frac{120}{7}$, and $c = 0$.

Step2: Find the time of maximum speed

For a quadratic function $y=ax^{2}+bx + c$, the $x$-coordinate (in our case $t$-coordinate) of the vertex is given by $t=-\frac{b}{2a}$. [t=-\frac{\frac{120}{7}}{2\times(-\frac{60}{49})}=-\frac{\frac{120}{7}}{-\frac{120}{49}}=\frac{120}{7}\times\frac{49}{120}=7]

Step3: Calculate the maximum speed

Substitute $t = 7$ into the function $s(t)$: [s(7)=-\frac{60}{49}\times7^{2}+\frac{120}{7}\times7=-\frac{60}{49}\times49 + 120=- 60+120 = 60]

Step4: Find when the car stops

The car stops when $s(t)=0$. So we set $-\frac{60}{49}t^{2}+\frac{120}{7}t = 0$. Factor out $t$: $t(-\frac{60}{49}t+\frac{120}{7})=0$. We have two solutions: $t = 0$ (corresponds to the start - when the car is parked) and $-\frac{60}{49}t+\frac{120}{7}=0$. Solve $-\frac{60}{49}t+\frac{120}{7}=0$ for $t$: [-\frac{60}{49}t=-\frac{120}{7}] [t=\frac{-\frac{120}{7}}{-\frac{60}{49}}=\frac{120}{7}\times\frac{49}{60}=14]

Answer:

The maximum speed is 60 mph. It takes 14 seconds for the car to come to a complete stop again.