Question

A park ranger driving on a back country road suddenly sees a deer in hisheadlights 20 m ahead. The ranger, who is driving at 11.4 m/s, immediatelyapplies the brakes and slows down with an acceleration of 3.80 m/s2. Howmuch distance is required for the ranger's vehicle to come to rest? Onlyenter the number, not the units.

Answer 1

17.1 m

Step-by-step explanation:

To know how much distance is required to get a final velocity of 0 m/s, we need to use the following equation:

`v^2_f=v^2_0+2a(\Delta x)`

Where vf is the final velocity, v0 is the initial velocity, a is the acceleration and Δx is the distance required. So, replacing vf by 0 m/s, v0 by 11.4 m/s, and a by -3.8 m/s² we get:

`\begin{gathered} 0^2=(11.4)^2+2(-3.8)(\Delta x) \\ 0=129.96-7.6(\Delta x) \end{gathered}`

Then, solving for Δx, we get:

`\begin{gathered} 7.6\Delta x=129.96 \\ \Delta x=(129.96)/(7.6) \\ \Delta x=17.1\text{ m} \end{gathered}`

Therefore, the vehicle required 17.1 m to come to rest.