Question

A park ranger driving on a back country road suddenly sees a deer in his headlights 20m ahead. The ranger, who is driving at 11.4 m/s, immediately applies the brakes andslows down with an acceleration of 3.80 m/s2. How much distance is required for theranger's vehicle to come to rest? Only enter the number, not the units,

Answer 1

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

`\Delta x=(v^2-v^2_0)/(2a)`

*Here v = 0 m/s is the initial speed of the ranger's vehicle

Substitute the values in the above expression as

`\begin{gathered} \Delta x=(0^2-(11.4)^2)/(2*(3.80)) \\ =17.1\text{ m} \end{gathered}`

The distance is required for the ranger's vehicle to come to rest is calculated as

`\begin{gathered} D=20-17.1 \\ =2.9\text{ m} \end{gathered}`

The stopping time is calculated as

`v=v_0+at`

Substitute the values in the above expression as

`\begin{gathered} 0=11.4+(-3.80)(t) \\ t=3.00\text{ s} \end{gathered}`