Week 3 Quiz1v3

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course Phy 121

9/14 11:20pm

Solving Uniform Acceleration Problems

Possible Combinations of Variables Direct Reasoning

Using Equations Problem

Possible Combinations of Variables

There are ten possible combinations of three of the the five variables v0, vf, a, Dt and Ds. These ten combinations are summarized in the table below:

1

v0

vf

a

2

v0

vf

dt

3

v0

vf

ds

4

v0

a

dt

5

v0

a

ds

*

6

v0

dt

ds

7

vf

a

dt

8

vf

a

ds

*

9

vf

dt

ds

10

a

dt

ds

If we know the three variables we can easily solve for the other two, using either direct reasoning or the equations of uniformly accelerated motion (the definitions of average velocity and acceleration, and the two equations derived from these by eliminating Dt and then eliminating vf).

Only two of these situations require equations for their solution; the rest can be solved by direct reasoning using the seven quantities v0, vf, a, Dt, Ds, Dv and vAve. These two situations, numbers 5 and 8 on the table, are indicated by the asterisks in the last column.

Direct Reasoning

We learn more physics by reasoning directly than by using equations. In direct reasoning we think about the meaning of each calculation and visualize each calculation.

When reasoning directly using v0, vf, `dv, vAve, `ds, `dt and a we use two known variables at a time to determine the value of an unknown variable, which then becomes known. Each step should be accompanied by visualization of the meaning of the calculation and by thinking of the meaning of the calculation. A 'flow diagram' is helpful here.

Using Equations

When using equations, we need to find the equation that contains the three known variables.

We solve that equation for the remaining, unknown, variable in that equation.

We obtain the value of the unknown variable by plugging in the values of the three known variables and simplifying.

At this point we know the values of four of the five variables.

Then any equation containing the fifth variable can be solved for this variable, and the values of the remaining variables plugged in to obtain the value of this final variable.

Problem

Do the following:

Make up a problem for situation # 3, and solve it using direct reasoning.

Accompany your solution with an explanation of the meaning of each step and with a flow diagram.

Then solve the same problem using the equations of uniformly accelerated motion.

Make up a problem for situation # 8, and solve it using the equations of uniformly accelerated motion.

#3 If a ball starts from rest and rolls down a track 12 cm long, and is determined to be traveling at at 8 cm/s at the 12 cm mark, how long did it take to travel the 12 cm distance and, assuming uniform acceleration, what was the average acceleration?

We know the v0 is 0 cm/s and the vf is 8cm/s and the `ds is 12cm, we can reason out the two remaining values. From v0 and vf we can determine the average velocity. 0cm/s added to 8cm/s is 8 cm/s. Dividing by 2 we get an average velocity of 4 cm/s. Now since we have a distance of 12 cm to cover and the ball travels at 4 cm every second, we can see that it will take 3 seconds to cover that distance. We can then reason that if it takes 3 seconds for the velocity to change from 0 cm/s to 8 cm/s then, the acceleration has to be 8/3 cm/s per second.

I have a handwritten flow diagram but I don't know how to create it in text so I will just have to describe it.

The first line has v0 vf and `ds

The second line has vAve with lines going back to v0 and vf

The third line has `dt with lines going back to vAve and `ds

The fourth line has ""a"" with lines going back to vAve and `dt.

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Acceleration is not determined by vAve and `dt, but by `dv and `dt.

`dv follows from v0 and vf, so can occur at the second level. Then connecting `dv to `dt will give us a.

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The lines going back mean that the value was reasoned from the two values connected by the lines going back.

???Is there a better way to present a flow diagram in text format???

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This is exactly the way I describe these diagrams. There might be a better way but I haven't found it.

Fortunately it only takes a minute to type out the description, and you aren't asked for a lot of descriptions of flow diagrams.

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Using eqations we can solve the #3 problem like this:

`ds = (v0 + vf) / 2 * `dt

12cm = (0cm/s + 8cm/s)/2 *dt

12cm = (8cm/s)/2 * `dt

12cm = 4cm/s * `dt Divide each side by 4cm/s

3s = `dt

vf = v0 + a * `dt

8cm/s = 0cm/s + a*3s We can remove the 0cm/s

8cm/s = a* 3s Divide each side by 3s

8/s cm/^2 =a

a = approximately 2.7 cm/s^2

#8 a car is rolling down a uniformly sloped grade acclerating at a rate of 2 meters/s^2 passes a mile marker. Ater travelling an additional 75 meters the car is travelling at a velocity of 20 m/s, how long did it take the car to travel the 75 meters and what was its initial velcotiy at the mile marker?

We'll use equations to figure out this one.

vf^2 = v0^2 + 2a * `ds

(20 m/s)^2 = v0^2 + 2(2m/s^2) * 75m

400m^2/s^2 = v0^2 + (4m/s^2 * 75m)

400m^2/s^2 = v0^2 + (300 m^2/s^2) Subtract (300 m^2/s^2) from both sides

100m^2/^2 = v0^2 Take the square root of both sides

10m/s = v0

Then we just need to find `dt

`ds = (v0 + vf) / 2 * `dt

75m = (10m/s + 20m/s)/2 * `dt

75m = (30m/s)/2 * `dt

75m = 15m/s * `dt Divide both sides by 15m/s

5 s = `dt

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&#This looks good. See my notes. Let me know if you have any questions. &#