Help please - extreemly stuck

I need to work out the following simultaneous linear equation:


a is constant acceleration, u is initial velocity, t is time

after times of 3 seconds and 4 seconds the distances travelled were 25mtr and 35mtr

determine the value of u and a

Please help I am real lost on this one.

I need to work out the following simultaneous linear equation:

s=ut+1/2at²

a is constant acceleration, u is initial velocity, t is time

after times of 3 seconds and 4 seconds the distances travelled were 25mtr

x.

Yes, I need to solve a pair of simultaneous equations. Unfortunatly I missed this section due to time off with an illness. If you could help with workings out it would be appreciated so I can understand it and complete the remaining questions I have.

So you can form 2 questions from this info;

$\displaystyle (25) = u(3) + \frac{1}{2}a(9) , \quad \text{call this equation}\: (1)$

and

$\displaystyle (35) = u(4) + \frac{1}{2}a(16), \quad \text{call this equation}\: (2)\:$

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$\displaystyle (1) \: \text{simplifies to;}\quad 25 = 3u + \frac{9}{2}a$

$\displaystyle (2) \: \text{simplifies to;}\quad 35 = 4u + 8a$

have you got his far?

Thanks Dylan, so is that the end /answer?
How do I find the values of u and a?
Sorry, this topic is totally new to me

Thanks Dylan, so is that the end /answer?
How do I find the values of u and a?
Sorry, this topic is totally new to me

Those last two are basic simultaneous equations, you should be able to solve that

Thanks Dylan, so is that the end /answer?
How do I find the values of u and a?
Sorry, this topic is totally new to me

Don't be sorry, its great that you're asking for help

We have two equations in front of us which are;

$\displaystyle (1) ; \quad 25 = 3u + \frac{9}{2}a$

$\displaystyle (2) ; \quad 35 = 4u + 8a$


I will multiply (1) by 2 as its never nice to work with fractions

$\displaystyle (1) ; \quad 50 = 6u + 9a$

$\displaystyle (2) ; \quad 35 = 4u + 8a$

This next step is very important, we have to remove a variable somehow, we have the choice to remove u or a, in this case it looks easier to remove u so heres what were going to do;


multiply (1) by 2 and multiply (2) by 3 (you'll see why soon). (1) and (2) then become;

$\displaystyle (1) ; \quad 100 = 12u + 18a$

$\displaystyle (2) ; \quad 105 = 12u + 24a$


Notice we've got the coefficient of u the same. So lets do $\displaystyle (2) - (1)$ (you can do (1)-(2) if you wish but it makes everything minus which is boke)

$\displaystyle (2)-(1); \quad 105-100 = 12u -12u +24a -18a$

$\displaystyle (2)-(1); \quad 5 = 6a \implies \boxed{a = \frac{5}{6}}$

We now have a value of a, so now all thats left to do is find the value of u, so well go back to (1) and (2) and choose one to sub our value of a into (again choose whichever looks easiest.)

Let's choose $\displaystyle (1) ; \quad 50 = 6u + 9a$

so subbing in our value for a we get;

$\displaystyle (1) ; \quad 50 = 6u + 9\left(\frac{5}{6}\right) \implies \boxed{u = \frac{85}{12}}$

This part is optional; sub your values for u and a into (2) to check if it works

$\displaystyle (2) ; \quad 35 = 4\left(\frac{85}{12}\right) + 8\left(\frac{5}{6}\right)$ which is correct, so were done

Edit: holy mother of god i spent about 30mins on this... my latex is so slow

Edit II: sorry for full solution btw but i felt OP would benefit more from seeing a worked example rather than small hints, but maybe that's not my decision to make

Edit III; T.I.L. \boxed

Dylan, that it brilliant thank you so much.
I understand it until where you get the value of U? How do you get that fraction?