Physics Chapter 3 questions 2

Question

Chapter 03, Problem 10


Interactive Learningware 3.1 reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 1.60 km due east, 0.927 km due south, and 1.08 km in a direction 63.7 ° north of west. The time required for this trip is 0.8161 h. Determine the magnitudes of the bird watcher's (a) displacement and (b) average velocity.




(a) Number

Units

kmkm/h


(b) Number

Units

kmkm/h

Answer 1

Hi Rae!

 

For this problem, the general mathematical technique (as opposed to drawing vectors), is to represent these motions as individual vectors, then

 

(a) break those vectors down into components,

(b) add the corresponding components of each vector, to get the components of the sum vector, then

(c) build the sum vector out of its components, and use it to answer the questions asked.

 

The first vector (let's call it A) has a magnitude of 1.63 km, and a direction of due east (which we can define as the positive x direction).  Because of this, it only has an "x" component; all of this vector's magnitude is in the x direction.  Therefore, we can write its x and y components as:

 

A_x = 1.60 km

A_y = 0 km

 

The second vector, B, is due south (which we can call the negative y direction).  As such, it only as a "y" component.  So we can write:

 

B_x = 0 km

B_y = -0.927 km

 

Finally, we have a vector C that is neither only in x nor only in y.  It is in both, and we need to determine its components to find out "how much" of its effect is in each direction.  I usually do this with the help of a sketch, which I unfortunately cannot draw here.  But with the given angle, we can say about the components of C that:

 

C_x , with respect to the given angle, is adjacent to the angle and in the negative direction

C_y , with respect to the given angle, is opposite and in the positive direction

 

These two facts about each component tell me what I need to know to calculate them:

 

C_x, being adjacent, has a magnitude that relates the overall magnitude of C through the cosine function:

 

C_x = C(cosθ) = (1.08 km)cos(63.7^o) = 0.48 km

 

Since C_x points in the negative direction, I have to make sure to affix a negative sign to it, so:

 

Cx = -0.48 km

 

 

I can go through the same procedure for C_y, which relates to the overall magnitude of C through sine, and which points in the positive direction.  This gives:

 

C_y = C(sinθ) = (1.08 km)(sin(63.7°)) = 0.97 km

 

C_y = 0.97 km  (since it is positive)

 

*********************************************************************************************

 

Now I have all the components of each vector, so I can add them together into the components of the resultant (R).  The nice thing about adding x and y components is that, because they are all along the same line, they add just like positive and negative numbers (unlike general vector addition where direction also has to be taken into account).  So:

 

R_x = A_x + B_x + C_x = 1.60 km + 0 km - 0.48 km = 1.12 km

R_y = A_y + B_y + C_y = 0 km - 0.927 km + 0.97 km = 0.043 km

 

*********************************************************************************************

 

The vector R is the overall displacement over the three motions, and we have to determine its magnitude.  To get the magnitude of a vector, given the magnitudes of its components, use the Pythagorean theorem:

 

R = √(R_x² + Ry²) = √((1.12 km)² + (0.043 km)²) = 1.121 km  (basically the same as the x component, since the y component is so small)

 

This is the magnitude of the displacement (unless I made a math error, which is always possible; but the steps are sound ones to take; I will check the math again later).

 

The average velocity is just the displacement divided by the total time.  So its magnitude is just (1.121 km/0.8161 hr) = 1.37 km/hr

 

I hope this helps!  Just let me know if you have any problems with or questions about the technique, or any other questions at all.