Answer: [tex]7.407 m/s^{2} [/tex]
Explanation:
In this question we are asked to find the cheeta's accelaration and we assume it is constant. Therefore, we can use the folowing equation:
[tex]V=V_{o}+at[/tex] (1)
Where:
[tex]V=80 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=22.22 m/s[/tex] is the cheeta's final speed
[tex]V_{o}= 0 m/s[/tex] is the cheeta's initial speed
[tex]t=3 s[/tex] is the time in which the cheeta passes from [tex]V_{o}[/tex] to [tex]V[/tex]
[tex]a[/tex] is the cheeta's acceleration
Isolating [tex]a[/tex] from (1):
[tex]a=\frac{V}{t}[/tex] (2)
[tex]a=\frac{22.22 m/s}{3 s}[/tex] (3)
Finally:
[tex]a+7.407 m/s^{2}[/tex]
