Help with a word problem in physics

S

staceyrho

Junior Member
Joined
Jan 12, 2007
Messages
76
I'm having a bit of trouble with a problem. I think i"m using the correct formula but want to be sure. Here is my problem.

Bob heads out into a lake at an angle of 47degrees with respect to the shore. If his boad is capable of speed of 2.7 m/s, how far from land will he be in 6 min and 52 s? Answer in units of m.

The formula I am tring to use is: Vx= vcos 47degrees
Vy= vsin47degrees

then one I've figured that I did

d=squareroot X^2=y^2 to get my time.

is this correct or do I need to be using one of the many other formulas to get my answer?
thanks
 
the distance from land is:
v[sin 47degrees] t
where v is the velocity of the boat, and t is the time of travel

2.7[sin47][6x60+52] m answer

the square root of [vy^2+vx^2] times t is the distance from the starting point.

Arthur
 
staceyrho said:
Bob heads out into a lake at an angle of 47degrees with respect to the shore. If his boad is capable of speed of 2.7 m/s, how far from land will he be in 6 min and 52 s? Answer in units of m.

The formula I am tring to use is: Vx= vcos 47degrees
Vy= vsin47degrees
then one I've figured that I did
d=squareroot X^2=y^2 to get my time.
is this correct or do I need to be using one of the many other formulas to get my answer?
Click to expand...
WHY are you trying to get "time"? It is GIVEN!

Bob travelled 6*60+52 = 412 seconds, at 2.7 m per second;
so distance travelled = 412 * 2.7 = 1112.4 m

If you "draw" that out, then drop a perpendicular from end of travel line to
the shore, you'll get a right triangle with hypotenuse = 1112.4;
the distance from shore will be the length of the side opposite the 47degree
angle (or the length of the perpendicular).

Can you finish it now?
 
Arthur I'm using your formula but I think I'm doing something wrong.
I did the 2.7sin47degrees(6x60+52) and I'm getting 762.9
then for the second part [vy^2+vx^2] times t am I taking
762.9^2+o^2(6x60+52)?
 
Hello, staceyrho!

Bob heads out into a lake at an angle of 47° with respect to the shore.
If his boat is capable of speed of 2.7 m/s,
how far from land will he be in 6 min and 52 s?
Answer in units of m.
Click to expand...

In 6 minutes, 52 seconds (412 seconds), he moves \(\displaystyle 412\,\times\,2.7\:=\:1112.4\) feet.

Did you make a sketch?
Code: 
                    *
                  * :
                *   :
       1112.4 *     :
            *       :d
          *         :
        * 47°       :
    - * - - - - - - + -

Can you find \(\displaystyle d\) now?

 
I think so, thanks for the help. Can you tell I really suck at this stuff. The bad part is that I have to take this class and it's way more than I planned for.
 
staceyrho said:
The bad part is that I have to take this class and it's way more than I planned for.
Click to expand...
Gee Stacey, if it's "MORE than you planned for", then ain't you lucky ? :roll:
 
I'm so unlucky with this stuff that I still cannot figure out what the answer is. I don't know what to do add, subtract multiply sin cos whatever. I'm clueless and so frusterated.
 
Stacy....I think that "unprepared" is a much more appropriate term than "unlucky."

You have been given much help with this problem....if, in particular, you can't proceed from Soroban's response, I think you may have been placed in a class you're not prepared for.

Maybe you should talk to your academic advisor. Perhaps there is a course you should take BEFORE the one you're currently enrolled in.
 
I finally got it. I just had to keep trying. thanks for all your help. I just need to work closely with someone who is better at this than myself.
Stacey
 
You must log in or register to reply here.
Top