Change in Velocity Homework: Solving the Vector Difference

[Jimbo57]

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?

 

[Whovian]

How are you using a negative vector?

 

[Jimbo57]

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?

 

[NemoReally]

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.

 

[cupid.callin]

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.

 

[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 \sqrt{a^2+ b^2}= 12.8}and, because &amp;quot;northeast&amp;quot; has equal parts &amp;quot;north&amp;quot; and &amp;quot;east&amp;quot;, a= b.&lt;br /&gt; &lt;br /&gt; Alternatively, you could do this &amp;quot;trigonometrically&amp;quot;. Represent each velocity vector as a line, &amp;quot;13.9 m/s east&amp;quot; of length 13.9 to the right, and &amp;quot;12.8 m/s northeast&amp;quot; of length 12.8 making an angle of 45 degrees with the first line. The &amp;quot;difference&amp;quot; is the third side of that triangle.

 

[Jimbo57]

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 \sqrt{a^2+ b^2}= 12.8}and, because &amp;quot;northeast&amp;quot; has equal parts &amp;quot;north&amp;quot; and &amp;quot;east&amp;quot;, a= b.&lt;br /&gt; &lt;br /&gt; Alternatively, you could do this &amp;quot;trigonometrically&amp;quot;. Represent each velocity vector as a line, &amp;quot;13.9 m/s east&amp;quot; of length 13.9 to the right, and &amp;quot;12.8 m/s northeast&amp;quot; of length 12.8 making an angle of 45 degrees with the first line. The &amp;quot;difference&amp;quot; is the third side of that triangle.

&lt;br /&gt; &lt;br /&gt; Ahh so &amp;quot;Northeast&amp;quot; represents an exact angle of 45 degrees? I had a hunch of this, but couldn&amp;#039;t find it anywhere. That helps a lot!&lt;br /&gt; &lt;br /&gt; Thanks everyone else for your help too!

 

[cupid.callin]

Ahh so "Northeast" represents an exact angle of 45 degrees?

yes NE means 45 angle with both north and east direction,

 

[Crotaphite]

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 \sqrt{a^2+ b^2}= 12.8}and, because &amp;quot;northeast&amp;quot; has equal parts &amp;quot;north&amp;quot; and &amp;quot;east&amp;quot;, a= b.&lt;br /&gt; &lt;br /&gt; Alternatively, you could do this &amp;quot;trigonometrically&amp;quot;. Represent each velocity vector as a line, &amp;quot;13.9 m/s east&amp;quot; of length 13.9 to the right, and &amp;quot;12.8 m/s northeast&amp;quot; of length 12.8 making an angle of 45 degrees with the first line. The &amp;quot;difference&amp;quot; is the third side of that triangle.

&lt;br /&gt; &lt;br /&gt; Something about the question is throwing me off... &lt;br /&gt; &lt;br /&gt; Using vector subtraction I get 62° (61.8°) North of East -(Xtotal =4.85 m/s East &amp;amp; Ytotal =9.05 m/s North)&lt;br /&gt; Using trigonometry I get 62° (61.8°) North of West&lt;br /&gt; &lt;br /&gt; I think the second answer is right (North of West), not sure though. &lt;br /&gt; Can anyone help explain what I&amp;#039;m doing wrong?