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

17.1

Explanation:

The distance ahead, of the deer when it is sighted by the park ranger, d = 20 m

The initial speed with which the ranger was driving, u = 11.4 m/s

The acceleration rate with which the ranger slows down, a = (-)3.80 m/s² (For a vehicle slowing down, the acceleration is negative)

The distance required for the ranger to come to rest, s = Required

The kinematic equation of motion that can be used to find the distance the ranger's vehicle travels before coming to rest (the distance 's'), is given as follows;

v² = u² + 2·a·s

∴ s = (v² - u²)/(2·a)

Where;

v = The final velocity = 0 m/s (the vehicle comes to rest (stops))

Plugging in the values for 'v', 'u', and 'a', gives;

s = (0² - 11.4²)/(2 × -3.8) = 17.1

The distance the required for the ranger's vehicle to com to rest, s = 17.1 (meters).

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