A car is moving with a speed of 32.0 m/s. The driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s2. How far does the car travel after the driver put on the brakes until it comes to a stop?

292 m

Explanation:

Step one:

given data

initial velocity velocity v= 32 m/s

final  velocity u= 0 m/s

acceleration a= 3.5m/s^2

Required

The distance covered upon the application of the brake

Step two:

we know that acceleration

a= v-u/t

t= v-u/a

t= (0-32)/3.5

t=32/3.5

t=9.14seconds

also, to find distance, we  use s=d/t,

rearrange as d=s*t  

d=32*9.14 s

d=292 m

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more