1.2−0.3×0.2g=0.2a
*M1
Resolve horizontally using
Newton's Second Law; 3 relevant
terms; allow sign errors; R=0.2 g
only.
a=3
A1
0.6=0.2 a only seen, allow with
BOD, but if 0.6 as friction being used as resultant force, this is
M0A0.
s3=0+21×3×32[=13.5]s2=0+21×3×22[=6]
DM1
For use of s=ut+21at2 (or a
complete method) to find a distance
at least once with u=0 and their
positive a and t=2 or t=3.
Distance =13.5−6=7.5 m
A1
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