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!
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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
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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