Answer: [tex]a=7.12\ \frac{ft}{s^2}[/tex]
Explanation:
For this exercise we need to use th following formula:
[tex]d=V_0t+\frac{1}{2}at^2[/tex]
Where [tex]d[/tex] is the distance, [tex]V_0[/tex] is the initial velocity, [tex]a[/tex] is the acceleration and [tex]t[/tex] is the time.
The first step is to convert from [tex]31 \frac{mi}{h}[/tex] to [tex]\frac{ft}{s}[/tex]. Since:
[tex]1\ mi=5,280\ ft\\\\1\ h=3,600\ s[/tex]
We get:
[tex](31 \frac{mi}{h})(\frac{5,280\ ft}{1\ mi})(\frac{1\ h}{3,600\ s})=45.466\ \frac{ft}{s}[/tex]
Knowing that:
[tex]V_0=45.466\ \frac{ft}{s}\\\\t=6\ s\\\\d=401\ ft[/tex]
We can substitute values into the formula and solve for "a":
[tex]401=(45.466)(6)+\frac{1}{2}a(6)^2\\\\401=272.796+18a\\\\401-272.796=18a\\\\a=\frac{128.204}{18}\\\\a=7.122\ \frac{ft}{s^2}[/tex]
Rounded to the nearest 100th place:
[tex]a=7.12\ \frac{ft}{s^2}[/tex]
