060911

Continued from 060906:  http://vhmthphy.vhcc.edu/ > Physics I > Physics Homepage > Prob Sets > Set 1

also Set 2

Attempt each problem before reading the solution.

Read the entire solution and take notes on everything you see, especially the things you don't understand.  Come to class with questions about anything you don't understand.

060911Quiz

1.  Write down the four equations of uniformly accelerated motion.

Be sure you can write these down from memory.  You should also know how the first two equations can be obtained from the 'triangle' diagram using the following three things:

• definition of velocity
• definition of acceleration
• fact that if v vs. t is linear over an interval, the ave velocity on that interval occurs at the midpoint of both the time and velocity interval

2.  If a ball rolling up an incline accelerates uniformly from velocity 34 cm/s to velocity 10 m/s over a period of 12 seconds, then what is its acceleration and how far does it travel?

Include in your answer a sketch of v vs. t and an explanation of how this sketch could be used to obtain the answers to the question.

Include also a ‘triangle’ diagram illustrating this situation and explain how you would use this diagram at each step to answer the question.

Definition of acceleration:  rate of change of velocity with respect to clock time.  So ave accel = `dv / `dt.

• `dv is change in velocity, which is vf - v0 = 10 m/s - 34 cm/s = 1000 cm / s - 34 cm/s = 966 cm/s
• dt is change in clock time, given as 12 seconds
• aAve = `dv / `dt = 966 cm/s / (12 s) = 85 cm/s^2.

Definition of velocity:  rate of change of position with respect to clock time.  So ave vel = `ds / `dt.

Thus `ds = vAve * `dt.

Acceleration is uniform so vAve = (vf + v0) / 2 = (1000 cm/s + 34 cm/s) / 2 = 517 cm/s.

Finally, then, `ds = vAve * `dt = 517 cm/s * 12 s = 6204 cm.

Comments: 

• Not everyone noticed the difference in the velocity units.  No big problem this time, but do be careful about units.
• On some versions of the quiz the change in velocity was negative, which would result in a negative `dv and a negative acceleration.

3.  Which of the equations of uniformly accelerated motion could you use to obtain additional information about the following situation:

An automobile accelerates down an incline, starting at 20 m/s and accelerating at .5 m/s^2, and continuing to accelerate at this rate through a displacement of 200 meters.

This situation specifies v0, which is 20 m/s, acceleration a = .5 m/s^2 (since accel is uniform aAve can be written as just a), and displacement `ds = 200 m.

The equations

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

`ds = v0 `dt + 1/2 a `dt^2

each contain these three variables.  The first can be solved for vf, obtaining

vf = +- sqrt( v0^2 + 2 a `ds)

while the second could be solved for `dt.  However the equation is quadratic in `dt, and for right now at least we're going to avoid the quadratic.  Rather than solving the second equation for `dt, we can use vf (found just above) along with v0 and a in the second equation vf = v0 + a `dt, which we can easily solve for `dt.