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Chap 3 Questions.. PLease HELP>>>

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Scholar
Karma Points: 306
Respect (90%):
Date Posted: 7/2/2008 6:42:22 PM  Status: Closed
Chap 3 Questions.. PLease HELP>>>
Course Textbook Chapter Problem
General Physics College Physics (8th) by Young N/A N/A
Question Details:
A river flows due south with a speed of 2.00 {\rm m}/{\rm s}. A man steers a motorboat across the river; his velocity relative to the water is 4.20 {\rm m}/{\rm s} due east. The river is 800 {\rm m} wide.

Part A:What is the magnitude of his velocity relative to the earth?
Part B: What is the direction of his velocity relative to the earth?
Part C: How much time is required for the man to cross the river?
Part D: How far south of his starting point will he reach the opposite bank?

I am stuck please help me out..
Bonus Point Alert! Earn +4 additional karma points for helping this annual member.

Answers:

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Mentor
Karma Points: 422
Date Posted: 7/2/2008 7:48:52 PM  Status: Live
Asker's Rating: Somewhat Helpful   
Response:
The boat is going  4m/s sideways AND  2 m/s downward at the same time.
A)  To find net velocity, use pythagorean theorem  v = √[(22) + (42)]  =   √20
 
B)  To find the direction as a vector, just use the components you were given:
              v = 4i - 2 j
C)  Time =   Distance / Velocity   using only the x components.  (No conversions necessary since all units are in meters and seconds.
               t =  800 m / (4.2 m/s)     =  190 sec
 
D)  Using the same equation as above,    Distance = Time * Velocity,    this time using only the y components since we're looking for distance downstream.
               y =  190 sec  * (2 m/sec)   =   380 m
12345687155's Comment:
Answers A and B are wrong!

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Scholar
Karma Points: 306
Date Posted: 7/2/2008 8:26:43 PM  Status: Live
Asker's Rating: N/A-Posted by Person Asking Question   
Response:
I still cant figure out how to do part A and B?  Anyone out there help??

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Oracle
Karma Points: 14,581
Date Posted: 7/2/2008 8:37:18 PM  Status: Live
Asker's Rating: Lifesaver   
Response:
velocity of water vw = 2.00 m/s (S)
velocity of man relative to water vm/w = 4.20 m/s (E)
a) velocity of man v = vector sum of vw and vm/w,
magnitude of v = √(vw2 + vm/w2) = 4.65 m/s
b) tanθ = 2.00/4.20 = 0.4762
θ = 25.5o
direction: south of east 25.5o
c) time t = 800 m/4.20 m/s = 190.5 s
d) d = 2.00 * t = 2.00 * 190.5 = 381 m
12345687155's Comment:
Thank I see how its done now.. Thanks..




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