What is the final velocity of a car that accelerated 10 m/s2 from rest and traveled 180m?

Question

the answer is 60 m/s

pls. tell me how come the answer is 60 m/s??

please use these equations:
vf+vi/2
d=vt
vf=vi+at
t=d/v
a=v/t

When I'm trying to answer that question, my answer is always: 360 m/s..

Answer 1

What is the first equation? Does it mean average speed?
i.e. average speed = (vf+vi)/2 ..........(1)
d = vt here v means average speed not the final speed vf

For average speed we have to use equation (1).
Here the car continiously acclerated from rest and comes to final speed vf.
Therefore, you can not use the equation d = vt.
You have to use equation d =( (vf+vi)/2)*t

Or vf+vi = 2d/t (you have given vf = vi + at)
or t = (vf-vi)/a



Therefore, (vf+vi)*(vf-vi) = 2ad
or vf*vf = vi*vi + 2ad

Now vi = 0
a = 10
d = 180

Therefore vf*vf = 0+2*10*180
or vf = sqrt(3600)
or vf = 60

i.e the answer is 60m/s

Answer 2

Your given formulae are quite unclear, I use this:
v^2 = u^2 + 2as
v=final velocity, u=initial velocity, s=displacement
since u = 0(initial at rest)
v^2 = 0^2 + 2(10)(180)
v^2 = 3600
v= 60 m/s

Answer 3

The 2 important formulae are:
distance,s=ut+1/2(at^2) & final velocity, v=u+at
u=intial velocity,a=acceleration,t=time
Here final velocity has to be found out but time isnot known for that.As distance is given time can be found out from that.
Here s=180mt,u=0(as it started from rest) & a=10m/s^2
that is, 180=0*t+1/2(10*t^2)
that is, 180=5*t^2
that is, t^2=36.therefore time taken,t =6sec
Now v=u+at=0+10*6=60m/s

Answer 4

s = 180m
a = 10m/s^2
u = 0 (from rest)

v^2 - u^2 = 2as
v^2 - 0 = 2(10)(180)

v = 60m/s

Answer 5

d = (vf + vi)t/2
vf - vi = at
t = (vf - vi)/a
d = (vf + vi)(vf - vi)/2a
vf^2 - vi^2 = 2ad
vi = 0
vf^2 = 2(10 m/s^2)(180 m)
vf^2 = 3,600 m^2/s^2
vf = 60 m/s