Trouble with Uniformly Accelerated Motion problem

[Trickstar]

So I acquired an old Physics textbook (Gioncoli Physics 2nd Edition) out of which I am attempting to learn classical mechanics from. It's in Algebra and not Calculus so I thought I could do it since I just completed Advanced Algebra 2 w/ Trigonometry this year. The first chapter is on Kinematics in 1 dimension. So far I understand everything and have been able to do all of the exercises up to an example using uniformly accelerated motion. It is worded as follows:

Suppose a planner is designing an airport for small planes. One kind of airplane that might use this airfield must reach a speed before takeoff of 200 km/h (55.6 m/s) and can accelerate at 12.0 m/s². If the runway is 100 m long, can this airplane reach the proper speed to take off?

My technique style of learning is to try and solve for the uniform equations myself from the standard instead of just copying them so I'll learn and the book is really confusing when it comes to explaining things. The book gives the solution and which equation it used, but I don't really understand how it got there. I'll first show the books solution then show my work.

Book's solution:

We use eq. 1-10c ([itex]v^2 = v_{0}^{2} + 2a(x - x_{0})[/itex]) with x₀ = 0, v₀ = 0, x = 100 m, and a = 12.0 m/s². Then
[itex]v^2 = 0 + 2(12.0 m/s^2)(100 m) = 2400 m^2/s^2[/itex]
[itex]v = \sqrt{2400 m^2/s^2} = 49.0 m/s[/itex]

Unfortunately, this length runway is not sufficient. By solving Eq. 1-10c for (x - x₀) you can determine how long a runway is needed for this plane.

My work:

I figured that my given variables were a = 12 m/s² and x = 100 m. After that I figured I was supposed to solve for v:
[itex]a = \frac{v-v_{0}}{t}[/itex]
solves to:
[itex]v = v_{0} + at[/itex]
and plugging in the values it would simplify to:
[itex]v = 12t[/itex]

After this, I plugged into the values and simplified the velocity formula:
[itex]v = \frac{100}{t}[/itex]
Next I balanced the formulas so that I could solve for t, since it was another variable I needed. So:

[itex]12t = \frac{100}{t}[/itex]
simplifies to:
[itex]t = \sqrt{\frac{100}{12}}[/itex]
so then I had the values for t, x, and a:
[itex]v = 100/\sqrt{\frac{100}{12}} \approx 34.641[/itex]

So there's my problem. I don't understand how my solution for v is so different from the books. Thanks in advance for any help. I'm assuming that in an actual physics course I would memorize the basic equations and I'm expected to know how to work out the uniformly accelerated motion equations on the spot, otherwise there would be too many equations to memorize.

 

[CAF123]

After this, I plugged into the values and simplified the velocity formula:
[itex]v = \frac{100}{t}[/itex]

The error is here. When is ##d=vt## applicable?

 

[Trickstar]

The error is here. When is ##d=vt## applicable?

Because ##v=\frac{x_{2}-x_{1}}{t}## and x₁ = 0 and x₂ = 100 so that's what it simplifies to. At least how I'm thinking of it that's how it is. How else am I supposed to go about this? Am I supposed to memorize the Uniformly Accelerated Motion equations?

 

[CAF123]

Because ##v=\frac{x_{2}-x_{1}}{t}## and x₁ = 0 and x₂ = 100 so that's what it simplifies to. At least how I'm thinking of it that's how it is. How else am I supposed to go about this? Am I supposed to memorize the Uniformly Accelerated Motion equations?

What does using the equation ##(x - x_0) = vt## imply about an objects acceleration?

 

[Trickstar]

I was trying to solve for v, so I figured that since I had x, I could try to use it to solve for v, then solve for t and then plug it back into the equation. If this wasn't the right way, what way should I have went? I'm looking into my book and by the looks of it I'm just supposed to memorize these. Is that what I have to do in an actual physics course? I'm planning on taking AP Physics B my senior year, and I want to already know it so I could ask my teacher the Calculus versions of everything (since I will be in AP Calc BC) so I could take the AP Physics C exam.

EDIT:
I'm trying to find the time by using x and a. The method I am using makes the most sense to me. I know your trying have me think about it and I'm looking back at the book to figure out how the equations are combined to form the one in the books solution. It combines:
v= v₀ + at
x= x₀ + vt
v= (v + v₀)/2

It's making a little sense to me now.

 

[STEMucator]

You cannot apply ##Δd = vΔt## unless you have constant velocity. That's why you're getting the wrong answer.

You're given all the right info to use ##v_{2}^{2} = v_{1}^{2} + 2aΔd##, so why not use it?

 

[Trickstar]

You cannot apply ##Δd = vΔt## unless you have constant velocity. That's why you're getting the wrong answer.

You're given all the right info to use ##v_{2}^{2} = v_{1}^{2} + 2aΔd##, so why not use it?

I'm trying to make sense as to how the equations combine to form that. I won't truly understand until I figure out how that equation was algebraically created.

EDIT: I finally figured it out:
##v = v_{0} + at##
##x = x_{0} + v_{avg}t##
##v_{avg} = \frac{v + v_{0}}{2}##

You plug the third equation into the second to form:
##x = x_{0} + (\frac{v + v_{0}}{2})t##

Then you plug in the acceleration in in the first equation to find v.
I'm going to try to solve the problem using this now.