Find initial velocity and acceleration of a bike

[Girn261]

Homework Statement

A power plant operator on a bicycle is pedaling at uniform acceleration of "a", when his initial velocity "u" m/s. After 6.2s, and 9.3s, the bicycle had moved 23.5m & 39.7m respectively. Find the value of "u" & "a".

Homework Equations

s=ut+1/2at^2

The Attempt at a Solution

I have the solution, just have issues understanding it. Here is part of the solution.
23.5 = u6.2+1/2(a)(6.2^2) -> 23.5=u6.2+19.22a(9.3)

39.7= u9.3+1/2(a)(9.3^2) -> 39.7=u9.3+43.245a(6.2)

;

218.55=57.66u+178.746a
246.14=57.66u+268a
etc.

My question is, why is the 9.3 & 6.2 multiple added at the end of the equation in the beginning? That I can't seem to understand.

 

[Merlin3189]

... Here is part of the solution.
23.5 = u6.2+1/2(a)(6.2^2) -> 23.5=u6.2+19.22a(9.3)

39.7= u9.3+1/2(a)(9.3^2) -> 39.7=u9.3+43.245a(6.2)

;

218.55=57.66u+178.746a
246.14=57.66u+268a
etc.

My question is, why is the 9.3 & 6.2 multiple added at the end of the equation in the beginning? That I can't seem to understand.

Well they aren't included in the calculation, so I thought they were simply intended as labels to show what the equations refer to. But of course they are the wrong way round!
So I think it's their way of showing the next stage of calculation, multiply the first equation by 9.3 and the second by 6.2. I used to do that, but separated them well and included something expressive like, "mult bs by 9.3" (though here I actually multiplied by 3 and 2 because 9.3 and 6.2 are common multiples of 3.1)

 

[Cutter Ketch]

The first and third line are the point. They constitute a pair of coupled equations. You have two equations in two unknowns. The rest is just algebra to solve the coupled equations. You may use any method you like to solve the coupled equations.

Now what they wrote on the second and fourth line is just wrong. I believe, as Merlin said, they intend to indicate that they are multiplying the entire second equation by 9.3 and the entire fourth equation by 6.2. That would be one proper approach to solving the simultaneous equations, and that is the result they write on lines 5 and 6. However, that isn't what they wrote. I don't think it's fair to call them labels. The mathematical expression is clear, and they just didn't write what they meant. It's a mistake, but a mistake they didn't propagate to the next step.

 

[Girn261]

Thanks for your help guys, I still don't understand why the two equations are multipled by 6.2 & 9.3. Again thanks your help in getting me to understand this. It's been a while since I was in college.

 

[Cutter Ketch]

Thanks for your help guys, I still don't understand why the two equations are multipled by 6.2 & 9.3. Again thanks your help in getting me to understand this. It's been a while since I was in college.

Ahh, now that we understand what they meant to do, your question now is why would they do that. Ok, let me explain. The first and third line are coupled equations. That is to say they both must be true for the very same values of a and u. You could rearrange either single quation to get a in terms of u or u in terms of a, but how do you get down to numerical values for a and u? For that you have to use the two equations together. There are a couple of ways. For example you could solve one equation for u and then substitute that result into the second equation. The resulting equation would not have u in it and could be solved for a. Whichever approach you use the idea is to use a combination of the two equations to eliminate one of the variables and then solve the single variable equation for the remaining variable. That numerical result can then be substituted into either of the original equations to find the value of the second variable.

What they did here is another method. We know logically that if we have two equalities
L1=R1 and L2=R2, then it must be true that
L1 - L2 = R1 - R2. Another logical truth is that if we multiply both sides by a constant an equation will still be true. That is to say, IF L = R, THEN C*L = C*R. You can use these two facts to solve coupled equations. As I said, the idea is to eliminate one of the variables. If one of the variables has the same coefficient in both equations, then when you subtract the two equations (as in fact 1 above), that variable will be eliminated. You can use fact 2 to make the coefficient of one of the variables the same in both equations. Just multiply each equation by the coefficient of that variable found in the other equation. That will assure that the modified coefficient in both equations will be the same: the product of the original two coefficients. Having made the coefficients the same, when you now subtract the two equations, that variable will be eliminated leaving an equation in a single variable which can be solved for that variable.

So that's why they multiplied each equation by the coefficient of u in the other equation.