The minimum distance required to stop a car moving at 30.0 mi/h is 48.0 ft. What is the minimum stopping distance for the same car moving at 56.0 mi/h, assuming the same rate of acceleration?
Thank you so much for the help.
i am really trying to understand the concepts?
Thank you so much for the help.
i am really trying to understand the concepts?

use vf^2=v0^2 + 2ad
vf=final velocity = 0
v0=initial velocity
a=acceleration
d=stopping distance
solving for d, we have
d= v0^2/2a
(since a is negative)
we see from this that if you hold the acceleration constant, the stopping distance varies as the square of the initial velocity
so the faster speed is 56/30 greater than the original speed, so the stopping distance will be
(56/30)^2 = 3.48 times greater
that means the stopping distance at the higher speed is 3.48 x 48 ft = 167ft
vf=final velocity = 0
v0=initial velocity
a=acceleration
d=stopping distance
solving for d, we have
d= v0^2/2a
(since a is negative)
we see from this that if you hold the acceleration constant, the stopping distance varies as the square of the initial velocity
so the faster speed is 56/30 greater than the original speed, so the stopping distance will be
(56/30)^2 = 3.48 times greater
that means the stopping distance at the higher speed is 3.48 x 48 ft = 167ft

equation of concern is 2ad=v2^2  v1^2
v2=0 so equation becomes
2ad=v1^2
solve for a
a =v1^2/2d
if we want to answer in feet need to convert speed from mph to ft/s
30 miles / hr = 30 * 5250 /3600 = 43.75 ft / s
a = 43.75^2/(2*48) = 19.94 ft / s^2
now use same equation but solve for d
2ad=v1^2
ad = v1^2/2a
change 56 mph to ft/s
56*5250/3600=81.67ft/s
d=v1^2/2a = 81.67^2/(2*19.94)
d=167.2 ft
v2=0 so equation becomes
2ad=v1^2
solve for a
a =v1^2/2d
if we want to answer in feet need to convert speed from mph to ft/s
30 miles / hr = 30 * 5250 /3600 = 43.75 ft / s
a = 43.75^2/(2*48) = 19.94 ft / s^2
now use same equation but solve for d
2ad=v1^2
ad = v1^2/2a
change 56 mph to ft/s
56*5250/3600=81.67ft/s
d=v1^2/2a = 81.67^2/(2*19.94)
d=167.2 ft