Vector Change in Velocity: What is the Resultant Velocity after a Right Turn?

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In summary, the problem involves finding the change in velocity of a car that is initially traveling due south at 60 km/h and then makes a right turn, resulting in a velocity of due west at 80 km/h. This change in velocity cannot be calculated by simply subtracting the two speeds, as velocity is a vector quantity and takes into account both magnitude and direction. Instead, the change in velocity can be found by using vector addition and finding the resultant velocity, which in this case would be -10 km/h, pointing southwest.
  • #1
henry2221
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Homework Statement


A car proceeding due south at 60 km/h (V1) makes a right turn, after which it is traveling due west at 80 km/h (V2). What is its change in velocity (V2-V1)?


Homework Equations


(V2-V1)


The Attempt at a Solution


I obviously did 80 km/h - 60 km/h = + 20 km/h. However this is wrong. I don't see why, but it is wrong. The instructor mentioned solving this geometrically or something like that, but I have no clue what they're talking about.
 
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  • #2
You are confusing velocity (a vector quantity) with speed (a scalar quantity). Its an easy trap to fall into as we use the two terms colloquially to mean the same thing. I assume your instructor wants you to compute the magnitude of the change in the velocity vector, so in a sense he has been caught in the same trap.

Can you compute v2-v1 as a vector and then take its magnitude?
 
  • #3
You need to read about vectors in your course materials. Velocity is a vector; it has both magnitude and direction. If we take east as the positive x direction, and north as the positive y direction, then the x component of V1 is 0, and the y component is -60 km/h. I could also specify V1 by its magnitude (60 km/h), and by the angle an arrow pointing in the direction of motion makes with the positive x-axis (90 degrees, or pi/2 radians).
 
  • #4
D H said:
You are confusing velocity (a vector quantity) with speed (a scalar quantity). Its an easy trap to fall into as we use the two terms colloquially to mean the same thing. I assume your instructor wants you to compute the magnitude of the change in the velocity vector, so in a sense he has been caught in the same trap.

Can you compute v2-v1 as a vector and then take its magnitude?


So basically it means,

(-80 km/h West) - (-60 km/h South) = -20 km/h

?
 
  • #5
First, why did you add those negative signs, and second, why did you throw away the directions?

I'm going to ask a different question: A car travels due south 60 km, turns to the west, and travels another 80 km. How far is the car from the starting point?

The answer is not 140km, nor 20km, nor anything like that. Don't throw away the directional information because that is one of the two things that make a vector a vector.
 
  • #6
D H said:
First, why did you add those negative signs, and second, why did you throw away the directions?

I'm going to ask a different question: A car travels due south 60 km, turns to the west, and travels another 80 km. How far is the car from the starting point?

The answer is not 140km, nor 20km, nor anything like that. Don't throw away the directional information because that is one of the two things that make a vector a vector.

I added those negative signs because in vectors south and west are considered negative in terms of magnitude? And I didn't throw away the directions? What did you mean? I mentioned the course of direction for each vector.

(-60 km due south)i + (-80km due west)j = -140 km due south west

?
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/1DKin/U1L1d.html
 
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  • #7
Are you just trying things at random now? You "mention" the course of direction but do nothing at all with it! Also, by the way, the units are NOT "km", they are "km/hr". YOU said that this was a vector problem. Why haven't you used vectors at all?

Draw a picture. Draw a vector "60 km/h" due south. In other words, pick some convenient scale (say a cm for each 10 km/h) and take "due south" to mean "down" on your paper. You have a line segment (arrow) 6 cm long going down. At the origin (beginning point) of that arrow, draw another "80 km/h due West". That would be a line segment (arrow) 8 cm long pointing to the left. Now draw the segment connecting the origin of the tips of the two arrows. That vector is their difference. You will need to give both its length (that's easy- think "3-4-5" right triangle) and the angle it makes (a bit harder- you will need trig functions).
 
  • #8
HallsofIvy said:
Also, by the way, the units are NOT "km", they are "km/hr".
Mea culpa! I thought Henry might be able to deal with the slightly different and more concrete problem of adding displacement vectors. My hope was that he would have readily answered my question in post#5. My plan was to then move him back to the original problem of adding velocity vectors. The best laid plans ...

Henry, please follow Halls' advice. Draw a picture.
 
  • #9
henry2221 said:

Homework Statement


A car proceeding due south at 60 km/h (V1) makes a right turn, after which it is traveling due west at 80 km/h (V2). What is its change in velocity (V2-V1)?


Homework Equations


(V2-V1)


The Attempt at a Solution


I obviously did 80 km/h - 60 km/h = + 20 km/h. However this is wrong. I don't see why, but it is wrong. The instructor mentioned solving this geometrically or something like that, but I have no clue what they're talking about.

I believed this might have been mentioned... You have to make a triangle with the vectors... Vectors are drawn head to tail... So your resulting Vector will point South West... You can use pythagorean theorem to find the resulting velocity... very difficult to explain vectors without drawing diagrams :(
 
  • #10
Why would I need to make a triangle when the problem is only asking for the change in velocity, it's not asking for the resultant. But i'll do that anyways. I'll post it tommorow.
 
  • #11
Because the change in velocity is (directly related to) the magnitude of the resultant :)
 
  • #12
CompuChip said:
Because the change in velocity is (directly related to) the magnitude of the resultant :)

You beat me to it. :)
 
  • #13
Would I use the avg velocity to find the change in velocity?

with the change in distance over time?

(-80 km) west - (-60 km) south / 2h - 0 h

thus

-20km/2h = -10 km/h due south west?

is this right?
 
  • #14
Why are you refusing to do what everyone here has asked you to do, which is draw a picture? This is a vector problem, you need to use vectors. You have'nt done that yet.

To answer your question, the result in post #13 is not right.
 
  • #15
D H said:
Why are you refusing to do what everyone here has asked you to do, which is draw a picture? This is a vector problem, you need to use vectors. You have'nt done that yet.

To answer your question, the result in post #13 is not right.

I drew it. So am I suppose to find the resultant? Or...?
 

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  • #16
That drawing is not quite right. It shows how to graphically compute V1+V2 (the vector along the hypotenuse). You want V2-V1. At least you have the magnitude correct now.
 

What is a vector?

A vector is a mathematical quantity that has both magnitude and direction. It is represented by an arrow with a specific length and direction.

What is a simple vector problem?

A simple vector problem is a mathematical problem that involves using vector concepts and operations to solve for a specific quantity or solve a real-world problem.

What are some common vector operations?

Some common vector operations include addition, subtraction, scalar multiplication, dot product, and cross product.

How can vectors be represented visually?

Vectors can be represented visually using arrows on a coordinate plane or graph. The length and direction of the arrow represent the magnitude and direction of the vector.

What are some real-world applications of vector problems?

Vector problems have many real-world applications, including calculating displacement, velocity, and acceleration in physics, navigation and direction in engineering and transportation, and force and motion in mechanics and sports.

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