Finding magnitude and direction of displacement.

In summary: and if they say the bear went 1563m west then turned 32 degrees north then headed 3348m in that direction, that's the angle that i use. so the cos(148) would go in front of the 1563 because that is the angle of the displacement.
  • #1
bling-bling
19
0

Homework Statement


In wandering, a grizzly bear makes a displacement of 1563m due west, followed by a displacement of 3348m in a direction 32 degrees NW. What are the (a) the magnitude and (b) the direction of the displacement needed for the bear to return to its starting point? Specify the direction relative to due east.

Homework Equations


R2 = A2 + B2 – 2ABcosθ


The Attempt at a Solution


a. magnitude= (1563^2)+ (3348^2)-2(1563)(3348)cos32=2185.53 m
b. SE
I don't know if i am right though because i don't understand what they mean by "due east"
 
Physics news on Phys.org
  • #2
welcome to pf!

hi bling-bling! welcome to pf! :smile:
bling-bling said:
… a displacement of 1563m due west, followed by a displacement of 3348m in a direction 32 degrees NW.

I don't know if i am right though because i don't understand what they mean by "due east"

i don't understand what you mean by "32 degrees NW" :confused:

"due east" means east …

for example, south-east would be 45° south of due east, and north would be 90° north of due east :wink:
 
  • #3
i have no idea what they mean either, but that's what it says in my assignment. do u want me to post a copy of the assignment so u can look at it?
 
  • #4
"32 degrees NW" is meaningless

they must mean either "32 degrees North of West" or "32 degrees West of North" :confused:
 
  • #5
i thought it was like the bear went 1563m west then turned 32 degrees NW and headed 3348 m in that direction?
 
  • #6
bling-bling said:
i thought it was like the bear went 1563m west then turned 32 degrees NW and headed 3348 m in that direction?

maybe :smile:

can't tell :redface:

the bear probably didn't know anyway :rolleyes:

i expect he just saw a sign saying "bear right" :biggrin:
 
  • #7
haha. but are my answers right?
 
  • #8
bling-bling said:
R2 = A2 + B2 – 2ABcosθ

a. magnitude= (1563^2)+ (3348^2)-2(1563)(3348)cos32=2185.53 m
b. SE

for the magnitude, shouldn't θ be greater than 90° ?

and to get the direction, you'll need to use the sine formula

(and I'm off to bed :zzz:)
 
  • #9
for a why would theta be greater than 90? and how do i use the sine formula for the direction?
 
  • #10
hi bling-bling;! :smile:

(just got up :zzz: …)
bling-bling said:
In wandering, a grizzly bear makes a displacement of 1563m due west, followed by a displacement of 3348m in a direction 32 degrees NW.

θ is the internal angle of the triangle, and that is greater than 90° …

the bear goes West, then he "bears" right to go North-West-ish, not North-East-ish

another way of looking at it is that if θ was exactly 90°, the it would be Pythagoras, a2 = b2 + c2

if you make the angle more than 90°, then that increases a2, so you need a plus, so -cosθ has to be positive, ie cosθ has to be negative, ie θ > 90°​

(and you use the sine formula sinA/a = sinB/b because you now know all three sides and one angle, and the sine formula is a lot quicker and easier to use than the cosine formula :biggrin:)
 
Last edited:
  • #11
so i use the sine formula to find the direction but what i got for magnitude, is that right?
 
  • #12
bling-bling said:
so i use the sine formula to find the direction but what i got for magnitude, is that right?

Your magnitude won't be correct since you used 32 degrees rather than [probably] 148 degrees.

Draw a diagram - [at least in your mind].

The first line is a long way due west - left.
The next line is up and to the left at 32 degrees.

Join the end of the second line to the start of the first [the starting point] and you get a triangle with a 148 degree angle and two very small angles.

You want that third side - and while it will stretch a long way in the direction "a few degrees North of due West", the answer you want, the return journey, is just as far, a few degrees South of East.

Peter
 
  • #13
I also got this question for my AP Summer Class Work. If you didn't find out the answer yet I think I've managed to get it myself.

I used the same Trig Ratio you used to get the answer just a different angle.

So my Equation looked something like this: X^2=(3348^2)+(1563^2)-(2*3348*1563)-Cos(148)

This gave me the answer of 1785m

Since the question didn't ask for an angle of the displacement I just said that the displacement was south of east.

Hopes this helps anyone else that has problems with this question.
 
  • #14
caleb did u get 1785m as the magnitude? and why did u put a negative sign infront of the cos(148) because that doesn't belong there.. right??
 
  • #15
because what i have as an example in my homework is that they use the angle that is given. for example, if they say that the bear turned 65 degrees, that's the angle they use in the formula, in my example.
 

What is displacement?

Displacement is the distance and direction between an object's starting position and its ending position. In other words, it is the overall change in an object's position.

How do you find the magnitude of displacement?

The magnitude of displacement can be found by calculating the distance between the starting and ending positions using a ruler or measuring tool. This distance is typically represented by a straight line.

How do you find the direction of displacement?

The direction of displacement can be determined by drawing an arrow from the starting position to the ending position. The arrow will point in the direction of the displacement.

What units are used to measure displacement?

Displacement is typically measured in units of distance, such as meters or kilometers. However, it can also be measured in units of time, such as seconds or minutes, if the object is moving at a constant speed.

Why is displacement important in science?

Displacement is an important concept in science because it helps us understand the motion of objects and how they change position over time. It is also a key factor in calculating other important quantities, such as velocity and acceleration.

Similar threads

Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
962
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
6K
  • Introductory Physics Homework Help
Replies
4
Views
6K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
5K
Back
Top