A car is slowing down. What is the acceleration if the speed BEFORE slowing down was 14.76 m/s and it comes to rest in 39.308m?
I would really appreciate the help ASAP!
Thanks!
I would really appreciate the help ASAP!
Thanks!

use the equation Vf^2=Vi^2+2a*x
Vf=final velocity (since it comes to rest, this is zero)
Vi= initial velocity
a=acceleration
x=displacement
0=14.76^2+2a*39.308
a= 2.77m/s^2
Vf=final velocity (since it comes to rest, this is zero)
Vi= initial velocity
a=acceleration
x=displacement
0=14.76^2+2a*39.308
a= 2.77m/s^2

v^2 = u^2 + 2ah
v = final velocity = 0 m/s
u = initial velocity = 14.76 m/s
a = acceleration
h = displacement = 39.308m
therefore,
a= (v^2 u^2)/(2h)
= (0  14.76^2)/(2*39.308)
=  2.77 m/s^2
hence, acceleration of 2.77 m/s^2 in opposite direction to velocity
v = final velocity = 0 m/s
u = initial velocity = 14.76 m/s
a = acceleration
h = displacement = 39.308m
therefore,
a= (v^2 u^2)/(2h)
= (0  14.76^2)/(2*39.308)
=  2.77 m/s^2
hence, acceleration of 2.77 m/s^2 in opposite direction to velocity