course PHY 231
For each situation state which of the five quantities v0, vf, `ds, `dt and a are given, and give the value of each. •A ball accelerates uniformly from 10 cm/s to 20 cm/s while traveling 45 cm.
We know initial velocity, final velocity, and distance. We need to know acceleration and time. First I calculate ‘dt.
‘ds = ((v0 + vf) / 2) * ‘dt
45 cm = ((10cm/sec + 20 cm / sec) / 2) * ‘dt
45 cm = ((30cm/sec) / 2) * ‘dt
45 cm = (15 cm / sec) * ‘dt
‘dt = 3 seconds
Now we need to find acceleration. I used the formula
vf^2 = v0^2 + 2a’ds
(20cm/sec)^2 = (10cm/sec)^2 + 2a(45cm)
400cm/sec = 100 cm/sec + 90a
400 cm/sec = 190a
a = 2.1 cm /sec^2
When you square cm/s you get cm^2 / s^2. The number 90 means 90 cm.
The correct equation would be
400cm^2/sec^2 = 100 cm^2/sec^2 + 90 cm * a.
In any case in your equation 100 cm/s is not a like term with 90 cm * a or with 90 * a. It is a like term with 400 cm/s.
In the correct equation 100 cm^2/s^2 is not a like term with 90 cm * a or with 90 * a. It is a like term with 400 cm^2/s^2 and your solution should have read
90 cm * a = 300 cm^2/s^2 so that
a = 300 (cm^2/s^2) / (90 cm) = 3.33... cm/s^2.
Better in the first place to have solved the symbolic equation for a then substituted (being careful, as always, with units).
answer/question/discussion:
• A ball accelerates uniformly at 10 cm/s^2 for 3 seconds, and at the end of this interval is moving at 50 cm/s.
We knoe vf, a, and ‘dt. We need to find v0 and ‘ds. First I solved for v0 using the following equation.
vf = v0 +a’dt
50 cm /sec = v0 + (10cm/sec^2) * 3sec
50 cm/sec = v0 + 30 cm/sec
20cm/sec = v0
Now I calculate ‘ds since I now have v0.
‘ds = v0*’dt + (1/2)(a)(‘dt)^2
‘ds = (20cm/sec) * (3 sec) + ½(10cm/sec^2)(3 sec)^2
‘ds = 60 cm + 1/2(90 cm)
‘ds = 60cm + 45 cm
‘ds = 105c,
answer/question/discussion:
• A ball travels 30 cm along an incline, starting from rest, while accelerating at 20 cm/s^2.
We know v0, a, and ‘ds. We need to find vf and ‘dt. First I calculate vf.
Vf^2 = v0^2 + 2a’ds
Vf^2 = 2(20cm/sec^2)(30cm)
Vf^2 = (40cm/sec^2)(30 cm)
Vf^2 = 1200cm^2/sec^2
vf = 34.64cm/sec
(I know that final velocity is not notated as VF, however I am having difficulty with Microsoft Word and I understand it’s noted as vf)
Now I find ‘dt.
Vf = v0 + a*’dt
34.64cm/sec = 20cm/sec^2 * ‘dt
‘dt = 1.73 sec
I’d like to note that with both equations I did not show v0 or v0^2/ This is because the initial velocity is 0. There was no need to show 0 in my work.
answer/question/discussion:
Then for each situation answer the following:
• Is it possible from this information to directly determine vAve?
No it is not. vAve = (v0 +vf)/2
I needed to find vf first..
answer/question/discussion:
• Is it possible to directly determine `dv?
No it is not. I need to find the final velocity before I can find a change.
answer/question/discussion:
You made some errors on the first problem so be sure to see my notes.
You did well on the second and third problems. Your use of the equations is good, but be careful.