AP C Linear Motion3,

**Problem 8**

This is a pretty touch problem. I began working on it this morning and had to stop and try again to find an easier solution using alternative kinematics equations than what I first tried. The key is recognizing that you can use a system of equations–multiple equations each describing a different aspect of the motion, with multiple unknowns. If you can find the right set of equations you can end up with a fairly simple solution. Here’s the one I found that worked best.

To stop a car, you require first a certain reaction time to begin braking. Then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 80.0 km/h, and 24.4 m when the initial speed is 47.8 km/h. These are the numbers from my problem, of course.

(a) What is your reaction time?

(b) Acceleration?

2 parts to each motion:

–Constant velocity before break is applied (during reaction time)

–During acceleration (slowing)

For constant velocity portion: v = x/t

For slowing motion: 2ax = v^2 – v0^2

Now, this is where it gets tricky. You will use 2ax etc. twice, once for each situation, but, and this is the big key here, you must substitute for x in that equation since the total x given in the problem is BOTH the constant speed distance traveled AND the slowing down distance traveled. As a result, you will need to subtract the constant velocity distance from the total stopping distance.

Basically you work will look like this:

Scenario #1:

2a(x1-v01*t) = V1^2 – vo1^2

Scenario #2:

2a(x2-v02*t) = V2^2 – vo2^2

X1 & X2 – given distances

V01 & V02 – given enitial velocities

t – reaction time (unknown)

a – acceleration (unknown)

V1 & V2 – final velocities, in this case both = 0

Notice the terms x – V0*t: These are you taking the distance given, and subtracting the pre-braking, constant velocity distance out of this formula, essentially leaving behind just the distance traveled during the negative acceleration.

Now, at this point you have 2 equations with 2 unknowns. Solve however you wish, but I prefer substitution. The algebra gets nasty. I actually had to redo my problem as I made one simple arithmetic mistake…not fun. Take it step by step.

Have fun!

**#16:**

This is one of my favorite problems on the assignment. It has elements of number 8 but is much easier. You know a couple pieces of information. In my version of the problem I know it falls from height h, and falls 0.49h in the last second. The easiest way to solve this is to not focus on the last second, but to focus on the first part of the fall instead. During whatever time that is (one second less than the total fall time) we know it falls 0.51h, the difference between 1 and whatever value of h you were given in your problem. This lets us set up two equations:

Whole fall: h = (1/2)gt^2

First part: 0.51h = (1/2)g(t-1)^2 <– this is the key. notice how we used the info from the problem, but we changed it up.

The reason I like to focus on the first part of the fall is because when Vo is zero, you get to cancel that variable. Makes the algebra MUCH easier.

So, we are left with 2 equations with 2 unknowns. Solve as you wish. I generally prefer substitution. In this case I simply take the first equation and plug in for h in the second, then used my graphing calculator to find the roots of the resulting quadratic to solve for time. But wait, you say! It’s quadratic so it has two solutions! You are absolutely correct, but one of them is impossible. I will eave it to you to figure out why, but I will say that one solution is outside the domain of the thing you are solving for.

**#15:**

Most common problem in dealing with this question is that the time given in your problem in part A–50ms–is 0.050s, not 0.50s. Remember the things on the stopwatch everyone calls milliseconds are in fact NOT MILLISECONDS. This is why a few of you have gotten answers with are simply 10x too big. Solving this question then is a simple application of a standard displacement function for constant acceleration, that kinematics equation that you know and love:

x=Vot + .5at^2, of course Vo = ?

Also note the units on the answer.

As for the rest of the problem you can either calculate each one individually, or recognize that because of the constant acceleration of gravity you are dealing with a quadratic relationship between distance fallen and time. That is to say, in two time the amount of time, the object will fall 4 times as far, or in three times the time, nine times as far, etc. Work it out if you are confused and look for a quadratic pattern in your answers.