Change in Velocity Homework: Solving the Vector Difference

In summary, a car's velocity changes from 13.9 m/s east to 12.8 m/s northeast. The change in velocity is 4 m/s in the northeast direction.
  • #1
Jimbo57
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0

Homework Statement


A car's velocity changes from 13.9 m/s east to 12.8 m/s northeast. What is the change in it's velocity?
It's a 4 marker.

Homework Equations





The Attempt at a Solution


Honestly, I don't really know how to start this. I know it's a vector difference and I've drawn the diagram which isn't too tough. The fact that I'm using a negative vector is throwing me off.

Could anyone nudge me in the right direction?
 
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  • #2
How are you using a negative vector?
 
  • #3
Whovian said:
How are you using a negative vector?

Vector subtraction?

v2-v1= v2+(-v1)

So does the 13.9 m/s here, not turn into a negative value?
 
  • #4
Jimbo57 said:

Homework Statement


A car's velocity changes from 13.9 m/s east to 12.8 m/s northeast. What is the change in it's velocity?
It's a 4 marker.

Homework Equations





The Attempt at a Solution


Honestly, I don't really know how to start this. I know it's a vector difference and I've drawn the diagram which isn't too tough. The fact that I'm using a negative vector is throwing me off.

Could anyone nudge me in the right direction?

It's the length and direction of the vector that goes from the first velocity to the second velocity. The difference vector isn't so much 'negative' as merely heading in a different direction.
 
  • #5
If you have studied about vectors and unit vectors then try finding unit vectors along north and north east.
Unit vectors have zero magnitude of 1 and a direction
So, vector(v) = magnitude(v) * direction (v)
Or, vector(v) = magnitude(v) * unit vector (v)

Just subtract the two vectors you get to get the final vector, and then find its magnitude.

PS: you wil always get a positive change as magnitude of a vector is always positive.
 
  • #6
You subtract. Subtract vectors. One way to do that is to write each vector in north and east "components"- ie < 13.9, 0> for the first vector and <a, b> for the second vector where [itex]\sqrt{a^2+ b^2}= 12.8}[itex] and, because "northeast" has equal parts "north" and "east", a= b.

Alternatively, you could do this "trigonometrically". Represent each velocity vector as a line, "13.9 m/s east" of length 13.9 to the right, and "12.8 m/s northeast" of length 12.8 making an angle of 45 degrees with the first line. The "difference" is the third side of that triangle.
 
  • #7
HallsofIvy said:
You subtract. Subtract vectors. One way to do that is to write each vector in north and east "components"- ie < 13.9, 0> for the first vector and <a, b> for the second vector where [itex]\sqrt{a^2+ b^2}= 12.8}[itex] and, because "northeast" has equal parts "north" and "east", a= b.

Alternatively, you could do this "trigonometrically". Represent each velocity vector as a line, "13.9 m/s east" of length 13.9 to the right, and "12.8 m/s northeast" of length 12.8 making an angle of 45 degrees with the first line. The "difference" is the third side of that triangle.

Ahh so "Northeast" represents an exact angle of 45 degrees? I had a hunch of this, but couldn't find it anywhere. That helps a lot!

Thanks everyone else for your help too!
 
  • #8
Jimbo57 said:
Ahh so "Northeast" represents an exact angle of 45 degrees?

yes NE means 45 angle with both north and east direction,
 
  • #9
Quote by HallsofIvy

You subtract. Subtract vectors. One way to do that is to write each vector in north and east "components"- ie < 13.9, 0> for the first vector and <a, b> for the second vector where [itex]\sqrt{a^2+ b^2}= 12.8}[itex] and, because "northeast" has equal parts "north" and "east", a= b.

Alternatively, you could do this "trigonometrically". Represent each velocity vector as a line, "13.9 m/s east" of length 13.9 to the right, and "12.8 m/s northeast" of length 12.8 making an angle of 45 degrees with the first line. The "difference" is the third side of that triangle.

Something about the question is throwing me off...

Using vector subtraction I get 62° (61.8°) North of East -(Xtotal =4.85 m/s East & Ytotal =9.05 m/s North)
Using trigonometry I get 62° (61.8°) North of West

I think the second answer is right (North of West), not sure though.
Can anyone help explain what I'm doing wrong?
 

1. What is change in velocity?

Change in velocity refers to the rate at which an object's speed and/or direction is changing. It is a vector quantity, meaning it has both magnitude and direction.

2. How is change in velocity calculated?

Change in velocity is calculated by taking the difference between the final velocity and initial velocity of an object. This is represented by the equation Δv = vf - vi, where Δv is the change in velocity, vf is the final velocity, and vi is the initial velocity.

3. What is the difference between average and instantaneous change in velocity?

Average change in velocity is the overall change in velocity over a given period of time, while instantaneous change in velocity is the change in velocity at a specific moment or instant in time.

4. How is the vector difference represented in change in velocity?

The vector difference in change in velocity is represented by the change in both magnitude and direction. This can be shown graphically using vector diagrams or mathematically using vector components.

5. What are some real-world applications of understanding change in velocity?

Understanding change in velocity is crucial in many scientific fields, including physics, engineering, and space exploration. It is used to calculate acceleration, determine the trajectory of objects, and predict the motion of planets and other celestial bodies.

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