Given data
*The speed of the ranger who is driving at 11.4 m/s
*The given acceleration is a = -3.80 m/s^2
The formula for the distance covered by the ranger's vehicle is given by the kinematic equation of motion as
[tex]\Delta x=\frac{v^2-v^2_0}{2a}[/tex]*Here v = 0 m/s is the initial speed of the ranger's vehicle
Substitute the values in the above expression as
[tex]\begin{gathered} \Delta x=\frac{0^2-(11.4)^2}{2\times(3.80)} \\ =17.1\text{ m} \end{gathered}[/tex]The distance is required for the ranger's vehicle to come to rest is calculated as
[tex]\begin{gathered} D=20-17.1 \\ =2.9\text{ m} \end{gathered}[/tex]The stopping time is calculated as
[tex]v=v_0+at[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} 0=11.4+(-3.80)(t) \\ t=3.00\text{ s} \end{gathered}[/tex]