v=u +at
s=ut+1/2at2
v2=u2+2as
s={(v+u)t}/2 where s-displacement
v-initial velocity
u-final velocity
a-acceleration/
decceleration
t-time
in this question we have:
t= 2s a=5m/s2 and, final velocity= om/s (after the car stops)
Therefore will use: s=ut + 1/2at2
s= 0 x 2 + 1/2 x 5 x 2(squared)
=0 + 1/2 x 5 x 4
=1/2 x 20
=10
therefore the distance covered = 10meter