You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ= 12.0°, that the cars were separated by distance d = 24.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was Vo= 18.0 m/s.
With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.60 (dry road surface) and (b) 0.10 (road surface covered with wet leaves)?

Explanation:

a Downward acceleration of car A along the slope

= g sinθ - μ g cosθ

= g ( sinθ - μ  cosθ)

= 9.8 ( sin 12 - .6 x cos 12 )

= 9.8 x ( .2079 - .5869 )

= - 3.714 m / s²

So there will be deceleration

v² = u² - 2 a s

= 18² - 2 x 3.714 x 24

= 324 - 178

= 146

v = 12 .08 m /s

b )

In the second case , kinetic friction changes

downward acceleration

= g ( sinθ - μ  cosθ)

= 9.8 ( sin12 - .1 x cos 12 )

9.8 ( .2079  -  .0978 )

= 1.079 m /s

there will be reduced acceleration

v² = u² - 2 a s

= 18² +2 x1.079 x 24

= 324 + 52

= 376

v = 19.4  m /s

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

it is very well related to a lot of things but with my limited knowledge i can confidently say that the answer is space.