Solving 2-D Vector Problems: Darryl's Distance & Displacement

[-Dragoon-]

Hey guys, it's me again!

Vectors have been really giving me a problem and turning me off from physics, but I hope I am getting the hang of it now.

Homework Statement

Darryl drives his load of tomatoes 14.0 km [E], 6.0 km [N], 12.0 km [ N 15° E], and then 2.0 km [N 65° E]. This takes him 42 minutes. Calculate Darryl's distance and displacement.
a) Calculate Darryl's distance and displacement. Draw a diagram and show your work.
b) Calculate Darryl's average speed and average velocity (record your answer in m/s).

Homework Equations

Magnitude * cosΘ = adjacent.
Magnitude * sinΘ = opposite.
C^2 = a^2 + b^2

The Attempt at a Solution

What I first did was break the 12.0 [ N 15° E] into components and found by using sine and cosine law that it was 14.7 km [E] and 3.1 km [N]. Then I broke down 2.0 km [N 65° E] into components and using the same method above I broke it down to 1 km [E] and 1.7 km [N]. Now that all my vectors are either parallel or anti-parallel, I added or subtracted where appropriate. I ended up with a total value of 29.7 km [E] and 10.8 km [N]. I drew these and then drew the resultant vector and found it's magnitude by using pythagorean theorem. Then I found the angle by using the tangent function. I ended up with 31.6 km [E 20° N] as the total displacement. Is this correct? Did I do it right How would I calculate Darryl's distance?

And for part b, to find the average speed would just be: total distance/total time and for average velocity: total displacement/total time, correct?

Thanks in advance!

 

[gneill]

Your general methodology looks to be okay. You might want to check out that first vector decomposition; the components of a vector should never be larger than the magnitude of the vector itself. You might also want to state your understanding of the directional notation (i.e., what does 12.0 [N 15° E] look like when you draw it? Where does the 15° angle sit?).

 

[-Dragoon-]

Your general methodology looks to be okay. You might want to check out that first vector decomposition; the components of a vector should never be larger than the magnitude of the vector itself. You might also want to state your understanding of the directional notation (i.e., what does 12.0 [N 15° E] look like when you draw it? Where does the 15° angle sit?).

Ah, I see where I made a mistake. 11.6 km rather than 14.7 km. The [N 15° E] is 15 degrees east of North. Why was there a problem in the direction of my resultant vector? That is the only thing I seem to struggle with, on how to determine the direction of the resultant vector. After I drew the resultant vector and made the triangle with the previous miscalculations ( 29.7 km [E] as the adjacent and 10.8 km [N]) then I would determine the direction of the resultant like that (which was pointing in the north-eastern direction). Am I doing this correctly? Thanks for all the help.

 

[gneill]

Draw a set of axes. Label the +Y direction N, the +X direction E, etc.

Your 12.0 [N 15° E] vector would be represented by a line of length 12 units and making an angle of 15° with the +Y axis. Note that the angle is *not* with respect to the x-axis. So consider carefully how you would determine the x and y components of that vector.

 

[-Dragoon-]

Draw a set of axes. Label the +Y direction N, the +X direction E, etc.

Your 12.0 [N 15° E] vector would be represented by a line of length 12 units and making an angle of 15° with the +Y axis. Note that the angle is *not* with respect to the x-axis. So consider carefully how you would determine the x and y components of that vector.

Okay, I think that is how I drew it in the first place being the angle was 15° east of North. I drew it with respect to the Y-axis. So, after I were to break down all the components, add and subtract when necessary and draw the resultant with the total distances traveled in each direction, and find the angle of the resultant it still be [N __° E], right?

 

[gneill]

Okay, I think that is how I drew it in the first place being the angle was 15° east of North. I drew it with respect to the Y-axis. So, after I were to break down all the components, add and subtract when necessary and draw the resultant with the total distances traveled in each direction, and find the angle of the resultant it still be [N __° E], right?

You'll have to decide which axis the resultant lies closest to, then specify how many degrees away from it it is.

 

[-Dragoon-]

You'll have to decide which axis the resultant lies closest to, then specify how many degrees away from it it is.

Alright, thanks for all the help. Appreciate it.