I got 1.667 for the accelaration.

well, if you multiply that by 2 you get the correct answer for acceleration at A (3.333...)

]]>With a = (v - u)/t, would t be equal to 6 (the total time from A to C) or 2 (The difference in time between AB and BC) or something different. If it's 2 then i got 5 for the accelaration which seems really wrong.

]]>luca-deltodesco, it appears you do have the correct answer, but I really don't understand your method. I fear you may have used techniques I haven't yet been taught, and hence, I doubt they'll need to be implemented in the exam. However, thanks for your input and well done for getting the correct answer.

Stardust, yep, that's it! I understand that, it's one of those answers you think, how the hell didn't I get that!? Haha. Fantastic! Thank you so much. And no, my M1 exam is on Tuesday! Do you get an extra day!?

.. well i dont remember any of the formulae, so i worked them out on the spot via integration (the top part), and then just applied it, and them simultaneous equation, anyways... i have my core 1 and 2 exams on tuesday

]]>Stardust, yep, that's it! I understand that, it's one of those answers you think, how the hell didn't I get that!? Haha. Fantastic! Thank you so much. And no, my M1 exam is on Tuesday! Do you get an extra day!?

]]>so in 4 seconds, the car travels 80m, and in 6s seconds, it travels 140m, since its acceleration is uniform, its velocity is linear, its displacement quadratic.

. now acceleration. its constant so, a(t) = k

and velocity is linear. taking as the integral, v(t) = kt + q

and distance is quadratic, taking as the integral, d(t) = 1/2kt^2 + qt + r

at t = 0, displacement is 0, so r = 0;

d(t) = 1/2kt^2 + qt

d(4) = 80 = 8k + 4q

d(6) = 140 = 18k + 6q

3.d(4) = 240 = 24k + 12q

2.d(6) = 280 = 36k + 12q

taking them away, 2.d(6)-3.d(4)

40 = 12k

k = 10/3

80 - 80/3 = 4q

q = 40/3

d(t) = (5/3)t^2 + (40/3)t

v(t) = (10/3)t + 40/3

a(t) = 10/3

t = 0, v(0) = 40/3, a(0) = 10/3

(wooh go me!) owned by a 15year old

]]>We know the acceleration is uniform, we will call this 'a'.

Looking at section AB:

S=80m

a=a

t=4 s

u: we do not know this, I will call this y

V:we do not know this, I will call this x

Looking at section AC:

S= 140

a = a

t =6

u= y

v we do not know

Substituting the values of AB into s = ut + 1/2 at² gives

80= 4y + 8a

dividing by 4 gives

y = 20 - 2a

This will be equation 1

Substituting the values of AC into s = ut + 1/2 at² gives

6y + 18a = 140

this will be equation 2

Substituting into 2 gives

6(20 - 2a) = 140

a = 20/6 = 3 1/3

Substituting back into equation 1

y=20 - 2x20/6

y= 13 1/3

Hope this helps. Are you doing the M1 exam on wednesday? I have the same text book, so found the question and checked my answers against the give answers.

]]>*"A, B and C are three points on a straight road such that AB = 80m and BC = 60m. A car travelling with uniform acceleration passes A, B and C at times t = 0 seconds, t = 4 seconds and t = 6 seconds respectively. Modelling the car as a particle find its acceleration and its velocity at A."*

This the only question in a Mechanics 1 book that I can't answer, or atleast got relatively close to answering. No matter what I try I'm always ending up with 3 unknowns and hence, am unable to apply the UVAST uniform acceleration equations. Someone, please help me with this! I beg of you! I would be very grateful for a worked answer so that I can understand it. Thanks a lot lads, you've helped me loads.

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