Problem 41 from Chapter 2
A problem where you choose to use algebra is going to have some givens and at least one thing to solve for. Assigning symbols to these entities is SO useful. But I’m taking it a step further. Add to your lexicon (which should already include Givens and Solving for What) the categories Hidden Given and Don’t Care. You will see what Hidden Given measn in the examples of these tutoring pages. A Don’t Care quantity is a physical quantity that can appear in algebra, but in the problem at hand, it is neither given nor asked for. Specifically identify it as Don’t Care. You will see why that is helpful.
For the Benz:
Given: v = 71.5 m/s
Hidden Givens:
- a = 0
- thev above is also v0. (v = v0 when a = 0)
Solving for: ∆x (Since “where are they?” is what is being asked.)
“Don’t Care” Category: This isn’t clear to me. This is because a = 0.
Whena = 0, the equations for the Benz become:
v2 – v02 = 2(0)∆x → v2 = v02
∆x = v0t + ½ (0)t2 → ∆x = v0t
v = v0 + (0)t → v = v0
∆x/t = ½ (v + v0) → ∆x/t = ½ (v0 + v0)
So when a = 0, the only pattern with any significance is ∆x = v0t. It falls out of equation 2 and 4 above and is nothing more than the area of the rectangle that lies underneath the Benz’s horizontal velocity versus time graph.
Therefore, solving for t is essential. XB is the position of Benz at the time that T-Bird returns to 71.5 m/s, and T is the time it takes for this to happen.
XB = v0T = (71.5 m/s)(T)
People who write expressions down precisely with good notation like above are people who figure out where to go with a problem. By writing down what’s above, one can see that T and XB are both unknown, and there is not enough info from Benz alone. Such a person now knows to look at T-Bird to get T. (That’s why it’s called a T-Bird.)
By the way, recognize the above expression for XBnames it as something different than xB. There is a huge distinction between XB and xB. By xB, I mean a changing location for the Benz, and it varies with time. By XB, I mean one particular value of xB, the particular value it takes when t = T. The importance of notation is for you to make the solution clear FOR YOURSELF so that you know where to go next. (Also, keep in mind that the XB and XB are the same thing here, because I decided that when all this started, the Benz was located at a place called x = 0.)
Where to Go Next:
For the T-BIRD:
T = Ttaken slowing + (5 s) + Ttaken returning again, organized thinking written down
I never know what to do when I start a problem. That is still true now after 28 years of doing these. The only thing I can do to get unstuck is to use good notation in my set-up.
For the same reason, I am hopelessly confused until I use the following display:
From the t = 0 to Ttaken slowing, the area under the graph is the area of a triangle. The value of this area is 250 m. In the language of areas of triangles, this takes the form:
Graphical Language:
(fill in the blanks)
Area = ½ (base)(height)
(250 m) = ½ (Ttaken slowing)( m/s) → Solve for Ttaken slowing
Alternate Method: in the language of “Given, Hidden Given, Solving for, Don’t Care, Choose Equation”, the above takes the following form:
Formulaic Language:
Given: v0 = 71.5 m/s, x = 250 m
Hidden Given: v = 0
Solving for: t (and it’s called Ttaken slowing if v is called 0, as it is.)
“Don’t Care” Category: a (a is not given and not asked for)
Choose the equation out of the following that doesn’t have a in it.
v2 – v02 = 2a∆x
∆x = v0t + ½ at2
v = v0 + at
∆x/t = ½ (v + v0)
Choice:
∆x/t = ½ (v + v0)
Substitution:
( )/Ttaken slowing = ½ [0 + ( )]
Ttaken slowing = ______
Using the Graphical Language and Formulaic Language above each succeeded in finding that Tslowing down equaled 6.99 s. If you were attentive to the algebra, you noticed that the two methods looked alike. They are the exact same thing. They have to be, because the equations only have validity due to the way that they were originally derived from graphs of motion like the one above.
Chapter 2 is not about learning to use the formulas. Chapter 2 is all about learning to communicate in different ways: graphical, pictorial, in words, and in the language of algebra. Success cannot be achieved without strength in all of these areas.
I’ll now leave it to you to figure out the other important time, the time called Ttaken returning, the time it takes the TBIRD to get back to 71.5 m/s. The distance of this speed up is given in the text of the problem. Find Ttaken returning using BOTH Graphical Language and Formulaic Language. You are responsible for both, so practice both. Prove through both methods that Ttaken returning = 9.79 s.
Final answer: Distance between cars = XB – XT
where XB = (71.5 m/s)(Ttakenslowing + (5 s) + Ttaken returning) and XT = 600 m
The answer in the textbook should confirm this method.
Why the Graphical Language?
The graph has reminded us to break down the problem into two separate sections, each of which has a different constant acceleration. (Actually 3 sections, counting when the TBIRD’s acceleration was 0 for 5 s.)
The 4 equations of the book may only be used when acceleration is constant. How are you going to really know when you need to break things up like in this problem? Well, a v versus t graph will always make that clear.
More Things to Watch Out for:
It doesn’t matter that the equations are given on the test. No equation in the world will reveal to someone the essential sub that v = 0 in this problem while the TBIRD comes to rest with constant a. A person can only know this by visualizing the problem and knowing what the definitions mean PHYSICALLY. Knowing when to put in v = 0 has nothing to do with math, and in this problem, failing to plug in v = 0 makes all formulas USELESS. So people who say that things are going to be easier because “he gives us the formulas” have no idea where the challenge lies.