t=1,

distance traveled=ut + 0.5at2

=0.5×1 +0.5x3x1sq

=2m

s=[(v-u)/2]t

2=[(v-3)/2]1

4=v-3

v=7m/s

t=2,

a=(v-u)/t

3=(v-7)/1

v=10m/s

t=3,

a=(v-u)/t

3=(v-10)/1

v=13m/s

t=4,

a=(v-u)/t

3=(v-13)/1

v=16m/s

t=5,

a=(v-u)/t

3=(v-16)/1

v=19m/s

t=6,

a=(v-u)/t

3=(v-19)/1

v=22m/s

However, I think there is something weird to this question. As you go up the hill, your velocity is suppose to slow down because you will encounter air resistance and also because the kinetic energy is converting to gravitational potential energy. This is unless you have disregarded it. Either that or I have done the question incorrectly somewhere….