Answer:
D. [tex]-360\,\frac{km}{h}[/tex]
Step-by-step explanation:
Physically speaking, average acceleration ([tex]\bar a[/tex]), measured in kilometers per square hour, is defined by following formula:
[tex]\bar a = \frac{v_{f}-v_{o}}{t_{f}-t_{o}}[/tex] (1)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speed, measured in kilometers per hour.
[tex]t_{o}[/tex], [tex]t_{f}[/tex] - Initial and final instant, measured in hours.
The initial and final instants are, respectively:
[tex]t_{o} = 5\,min\times \frac{1\,h}{60\,min}[/tex]
[tex]t_{o} = 0.083\,h[/tex]
[tex]t_{f} = 9\,min \times \frac{1\,h}{60\,min}[/tex]
[tex]t_{f} = 0.15\,h[/tex]
If we know that [tex]t_{o} = 0.083\,h[/tex], [tex]v_{o} = 60\,\frac{km}{h}[/tex], [tex]t_{f} = 0.15\,h[/tex] and [tex]v_{f} = 36\,\frac{km}{h}[/tex], then the average acceleration of Kelly is:
[tex]\bar a = \frac{36\,\frac{km}{h}-60\,\frac{km}{h}}{0.15\,h-0.083\,h}[/tex]
[tex]\bar a = -358.209\,\frac{km}{h^{2}}[/tex]
The correct answer is D.
