S5=at522=4⋅522=50mS_5=\frac{at_5^2}{2}=\frac{4\cdot5^2}{2}=50 mS5=2at52=24⋅52=50m
S4=at422=4⋅422=32mS_4=\frac{at_4^2}{2}=\frac{4\cdot4^2}{2}=32mS4=2at42=24⋅42=32m
Distance covered in 5th5^{th}5th second
S5−S4=50−32=18mS_5-S_4=50-32=18mS5−S4=50−32=18m Answer
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