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2 Different acceleration equations - when to use them ?

a=(v-u)/t

V^2 = U^2 +2aS ----> a = (v^2-u^2)/2S

I've done several past papers and whenever I choose to use one of them, it uses the other.

It seems the longer one (second one) gives an answer that is exactly half of the 1st one, Are there different scenarios that each one is unique to?
Original post by Walex12108
a=(v-u)/t

V^2 = U^2 +2aS ----> a = (v^2-u^2)/2S

I've done several past papers and whenever I choose to use one of them, it uses the other.

It seems the longer one (second one) gives an answer that is exactly half of the 1st one, Are there different scenarios that each one is unique to?


Are you confusing t = time (s) seconds in the first with s = distance (metres) in the second?
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Walex12108
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Original post by uberteknik
Are you confusing t = time (s) seconds in the first with s = distance (metres) in the second?


not that I'm aware of, here are two examples:

If you apply both methods to each question you get a different answer

to be fair before the first question it asked you to define acceleration; and then the correct method is to use the method the definition gives, but im still unsure if I wasn't prompted
(edited 8 years ago)
Question 1.JPG19.5KB
Question 1 MS.JPG6.6KB
Question 2.JPG22.3KB
Question 2 MS.JPG5.6KB
Original post by Walex12108
not that I'm aware of, here are two examples:

If you apply both methods to each question you get a different answer

to be fair before the first question it asked you to define acceleration; and then the correct method is to use the method the definition gives, but im still unsure if I wasn't prompted


how are you applying v2=u2+2as to the supertanker question?
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Walex12108
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Original post by Joinedup
how are you applying v2=u2+2as to the supertanker question?


I think I know why I cant use it...
I calculated the distance the supertanker moves at the 6m/s for 2400 seconds (14400m)
Which i subbed in for s and solved for 'a' - this isnt right because it is decelerating; it wont have the 6m/s for 2400 seconds instead this will decrease over time. ? is this correct.
its in the question after that it asks to calculate the distance the tanker travels.

Thanks for the help.
you cant use v^2=U^2 +2as
(edited 8 years ago)
Original post by Walex12108

I calculated the distance the supertanker moves at the 6m/s for 2400 seconds (14400m)


The tanker is decelerating and not moving at a constant velocity.

The tanker has travelled 7200 metres (not 14400 m) during the 2400 seconds deceleration time.

S=ut+12at2S = ut + \frac{1}{2}at^2

S=(6 x 2400) + 12(−6) x 24002S = (6 \mathrm{ \ x \ }2400) \ + \ \frac{1}{2}(-6) \mathrm{\ x \ }2400^2

S=14400−7200 = 7200 mS = 14400 - 7200 \ = \ 7200 \mathrm{ \ m}

Now substitute s = 7200 into the second equation and viola!
(edited 8 years ago)

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