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#1 2006-06-04 06:37:20

Tredici
Member
Registered: 2005-12-12
Posts: 28

Mechanics 1 - Kinematics Problem

question5fk.png

"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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#2 2006-06-04 08:53:54

stardust
Member
Registered: 2006-04-09
Posts: 48

Re: Mechanics 1 - Kinematics Problem

I think I have worked out how to do this.

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.

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#3 2006-06-04 09:03:28

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Mechanics 1 - Kinematics Problem

well i havnt started on mechanics yet, but lets see if i can do it;

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


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Last edited by luca-deltodesco (2006-06-04 09:07:16)


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#4 2006-06-04 09:31:37

Tredici
Member
Registered: 2005-12-12
Posts: 28

Re: Mechanics 1 - Kinematics Problem

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! sad Do you get an extra day!? sad

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#5 2006-06-04 09:37:12

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Mechanics 1 - Kinematics Problem

Tredici wrote:

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! sad Do you get an extra day!? sad

.. 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 smile


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#6 2006-06-04 18:04:12

stardust
Member
Registered: 2006-04-09
Posts: 48

Re: Mechanics 1 - Kinematics Problem

The exam is on Tuesday I was getting mixed up with chemistry I have on Wednesday. Ah too many exams

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#7 2006-06-05 19:38:48

Zmurf
Member
Registered: 2005-07-31
Posts: 49

Re: Mechanics 1 - Kinematics Problem

I got 1.667 for the accelaration.

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.


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#8 2006-06-05 21:33:08

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Mechanics 1 - Kinematics Problem

Zmurf wrote:

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...)


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