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Need some physics help

F

Finns14

Golden Member
I just need some help with a homework question this is the problem I have left if any one can explain to me how its done that would be great.

P.S. This is my professor's first class he ever taught



A bird watcher meanders through the woods, walking 1.92 km due east, 1.20 km due south, and 4.09 km in a direction 33.2 degrees north of west. The time required for this trip is 1.606 h. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity.


 
Draw out the displacement vectors along with their equations, the third one is a little annoying. Add the vectors together, find magnitude. That would be |d| = sqrt ( x^2 + y^2 + z^2 ). Take the displacement vector, divide by 1.606 to find your velocity vector. Then repeat the math to find the magnitude of that
 
Do you understand the basic concept of vectors? ( not being patronizing) oh ignore the z component of my last message since this isn't 3d space per se
 
Wait I don't think thats right because displacement means distance from the starting point it seems like you are solving for distance can any one tell me who is right?
 
Originally posted by: Finns14
Wait I don't think thats right because displacement means distance from the starting point it seems like you are solving for distance can any one tell me who is right?
Click to expand...

Distance traveled would be summing the magnitudes each individual vector. displacement is the magnitude of the resultant vector. Yeah? By adding the 3 vectors together you basically generate a "as the crow flies" line. To find the length of that line, you use the magnitude equation
 
Displacement is how far away you are from where you originally started.
Average velocity is the change in displacement divided by time.

So, your answer for (b) is going to be your answer for (a) divided by 1.606 h.

The only difficult part of this question is breaking the 4.09km vector into two perpendicular vectors. In fact, throughout physics, you'll normally want to break vectors into perpendicular components. So, you're going to use simple trigonometry. Draw a triangle, with the 33.2 degree angle. The hypotenuse is 4.09km. Find the other two sides. One of those sides will be the "north" component of that vector. The other side will be the "west" component of that vector.

add/subtract the east/west vectors, also add/subtract the north/south vectors.
(i.e. if you have a vector 5 km north and a vector 7 km south, your net in the n/s direction is 2 km south - you can do your own math on this problem.)

Now, draw a triangle with these two answers as legs of a right triangle. The hypotenuse will be your displacement.
 
Thank you Doc I think you cleared things up some, I get to try the problem and they tell me wether I am right or not and I will let you know. Also how does the angle apply to this problem when find the displacment.
 
Originally posted by: Finns14
Also how do you break down the northwest direction into a separate north and west value to add/subtract
Click to expand...
Use the cosine and sine of the 33.2 degree angle to find the component north and the one west.
 
Thanks sounds like I am getting closer how do I use sin and cos for distance? I know how to use the for velocity but not distance. Thanks so much.
 
Originally posted by: Finns14
Thanks sounds like I am getting closer how do I use sin and cos for distance? I know how to use the for velocity but not distance. Thanks so much.
Click to expand...
It's exactly the same as you would use it for velocity. All you're doing is finding the components of a vector. It doesn't matter if the vector represents distance, velocity, acceleration, etc.
 
well but when I used it for velocity I multiplied the sin or cos of the angle by the intial velocity. So in this case would I multiple the total distance by the sin and cos respectively?
 
Originally posted by: Finns14
well but when I used it for velocity I multiplied the sin or cos of the angle by the intial velocity. So in this case would I multiple the total distance by the sin and cos respectively?
Click to expand...
Yes, you would multiply the total distance by sine and cosine to find each component. You're doing the exact same thing you did for velocity - using the magnitude and angle to find each component. The only difference is what you call the magnitude - initial velocity vs. total distance. What you have to be careful about is using the sines and cosines correctly to get the right component depending on where the angle is with respect to the coordinate axis.
 
then the answer for (a) would be the hypotnuse(spelling) of the new triangle I get after subtracting the distances?
 
Did people forget that displacement is distance from your starting point?

You need to find that distance between his starting and ending point (a straight line). Divide that by the total time for his velocity.
 
A better way of visualizing what I mean is a car on a race track.

The car is going 80 mph on a circular track. It circles the track once.

The distance traveled is 2*pi*R
The displacement is 0 (since it ends up exactly where it started)
The average speed is 80 mph
The velocity is 0 (displacement over time)
 
Yes but its a matter of finding that distance from the starting point the NW distance is long than the hyptnuse and then how do you calculate that distance.


The car comparison didn't help
 
get some graph paper and draw yout the vectors. the final distance will be the vector sum of each step. this will give you the displacement. don't think of displacement as the distacnce travelled, think of it as how far it is from start to finish. as for average velocity, find the total distance travelled (which is the sum of the magnitudes of the individual displacement vectors) and divide by total time, which is give.

i hope that helps move you towards the end a little faster 😛
 
I don't know how to separate the north west vector into north and west this is my problem I understand that its the distance from the start but how do I find that distance.
 
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