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).

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.