Calculating distance but unsure of why I need to divide by two in working out

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HeeBeeJeeBee187
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#1
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#1
Acceleration= 1.5m/s^2
change in velocity= 18^2-9^2=243m/s
time taken=6seconds
distance= 81m

Working out is as follows
18^2 − 9^2= 2 × 1.5 × s
s = 18^2− 9^2 / (2 × 1.5)
can someone please explain the working out for me
thanks
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Joinedup
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#2
Report 5 months ago
#2
(Original post by HeeBeeJeeBee187)
Acceleration= 1.5m/s^2
change in velocity= 18^2-9^2=243m/s
time taken=6seconds
distance= 81m

Working out is as follows
18^2 − 9^2= 2 × 1.5 × s
s = 18^2− 9^2 / (2 × 1.5)
can someone please explain the working out for me
thanks
Could you post the whole question, those numbers don't seem to jive.

A change in velocity of 243 m/s in 6 seconds means the acceleration can't be 1.5 m/s/s
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HeeBeeJeeBee187
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#3
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#3
sure gimme a sec
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HeeBeeJeeBee187
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#4
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#4
https://pmt.physicsandmathstutor.com...y%201%20QP.pdf
Go to Q1) (f)
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HeeBeeJeeBee187
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#5
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#5
https://pmt.physicsandmathstutor.com...y%201%20MS.pdf
here is the markscheme
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HeeBeeJeeBee187
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#6
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#6
(Original post by Joinedup)
Could you post the whole question, those numbers don't seem to jive.

A change in velocity of 243 m/s in 6 seconds means the acceleration can't be 1.5 m/s/s
as you can see I am very confused XD posted the exam paper
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HeeBeeJeeBee187
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#7
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#7
(Original post by HeeBeeJeeBee187)
Acceleration= 1.5m/s^2
change in velocity= 18^2-9^2=243m/s
time taken=6seconds
distance= 81m

Working out is as follows
18^2 − 9^2= 2 × 1.5 × s
s = 18^2− 9^2 / (2 × 1.5)
can someone please explain the working out for me
thanks
doesn't matter I figured it out V^2-U^2=2as
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