100 point question!! A driver begins to brake when her car is traveling at 15.0 m/s, and the car comes to a stop 4.0 s later. How much farther does the car go after the driver begins to brake, assuming constant acceleration?

30 m

Explanation:

Let's list out the known variables that we are given in this question.

v₀ = 15 m/sv = 0 m/s t = 4 s

We are trying to solve for displacement in the x-direction.

Δx = ?

Find which constant acceleration equation contains all 4 of these variables.

Plug the known values into the equation and solve for delta x.

 

The car goes 30 m after the driver begins to brake.

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

One answer could be erosion