Relative Velocities Cont'd

So on with my second scribe post. ["Coincidentally" I was chosen by both previous scribes from Calculus and Physics to be the next scribe... I know, right?] I'll do my best as I kind of hurt my back in football practice and can barely sit right now, as I should be laying down in bed right now >=/.


We started off correcting the worksheet from the previous class. I'll just post the answers and a little summary for questions that need elaboration, because honestly, my back is killing me, but I was pretty sure everyone was in class, so...

NOTE: Judging by the number of people [including me] getting different answers, it is important to DRAW THE DIAGRAM CORRECTLY. So read the word problem carefully.

Answers:

4)
26.5 Degrees East of North @ 167.7 km/h [Answer may be round up]
-This was accomplished by correctly sketching the diagram, calculating the angle using the Tangent law and the Pythagoras theorem.

5)
Max Speed: 5m/s
Min Speed : 1m/s
Max Speed direction : With the current
Min Speed diraction : Against the current

6)
57 Degrees North of West @ 17m/s. [Note: the diagram is not a right angle triangle.]

7)
31.9 km/h
-If your not a "math" type person, you can just draw the diagram to scale (1cm = 10 km) and then from there use a protractor and trigonometric laws to find the angle to make a full triangle and then just measure the unmeasured line.

Following that, we were issued three more questions by Ms.K, which were questions 39, 40, and 41, from the second worksheet she handed to us yesterday.

39)
A) Use the sine law to find the answer which is; 13.6 degrees east of south.
B) Use the Pythagoras theorem to find the missing side to find an answer of 33 km/h.
C) Use the time equation (t = d/v) to find the time [21/33] = 0.64 h. To convert to seconds, multiply by 3600 seconds, and divide by one hour to find that there are 2304 seconds in that instance.

40) For this question we have to find which route is shorter, so we draw two diagrams complying with the word problem. (sorry for no diagrams, I can't stay on too long as my back hurts just to sit)
So for the first diagram, in which you try to swim directly across. t = d/v -> 1000m/ 2.2m/s = 454s.

For the second diagram in which you swim against the current, first we have to find the missing side. So we just use the Pythagoras theorem to find the missing side (c^2 - a^2 = b^2) and we get 1.51 m/s. From there we can find out how long it would take. t=d/v -> 1000m / 1.51m/s = 662 seconds.

So we find that swimming directly across would save us 208 seconds of time.

41)22 degrees north of east @16.2 km/h

After that, she gave us a sheet to read over. I did. Did you?

That's all for today folks. Sorry I couldn't elaborate on the questions more but honestly, my back's gonna give out. Hopefully I feel better tomorrow x( Now for an ice bath...

Oh and before I forget...

The next scribe is ERICT