Homework Help: Trying to Understand Accel Vectors

1. Sep 11, 2010

This is a sample problem in the text that I am having problems comprehending.

1. The problem statement, all variables and given/known data

A car is traveling northwest at 9.00 m/s. Eight seconds later it has rounded a corner and is headed north at 15.00 m/s. What are its magnitude and direction of its average acceleration during those eight seconds?

2. Relevant equations

∆v = V_f - V_i (vectors)

3. The attempt at a solution

The problem shows its strategy, but wants the student to do the calculations for each step.

Step 1: Write out V_i. Vector V_i=9m/s = -6.36m/s xhat + 6.36m/s yhat. I'm OK here. I know how to get velocity vectors and I worked this out correctly.

Step 2: Write out V_f. Vector V_f = 15m/s. I guess this is just a given. OK ....

Step 3: Calculate ∆V. ∆V= 6.36m/s xhat + 8.64m/s yhat. I don't know why the xhat is positive all of a sudden, but I understand that a turn due north adds zero to the xhat. I have no idea where the number for yhat is coming from.

There's a step 4 and 5, but if I can understand this part, I think I can work those out. Thanks in advance.

On a side note, how the heck can I type the symbols for hat, vectors, subscripts, etc.?

2. Sep 11, 2010

stevebac

I think the reason that the sign of the x component changes to positive is due to the fact that when you subtract the two vectors v_f and v_i, the signs of v_i changes so the x component becomes positive (becuase it was previously negative) and the y component becomes negative. So, since v_f has no x component in the first place, the resulting vector has the positive x component of the vector -v_i. The number for the y component appears to simply be 15 - 6.36 = 8.64, which is the result of subtracting the y components of v_f and v_i.

3. Sep 11, 2010

PhanthomJay

Subtracting vector A from vector B is the same as adding vector B to the negative of vector A, tht is B - A = B + (-A). The negative of a a vector is a vector pointing in the opposite direction.

(Edit: which is essentially what stevebac said).

4. Sep 11, 2010

Working on it now .....

5. Sep 11, 2010

OK, I'm still hopelessly lost, but here's what I've done to try and find the vectors:

Vector A = Ax+Ay = 9m/s
Ax = -6.36m/s
Ay = 6.36m/s

Vector B = Bx+By
Bx = 0
By = 15

Vector C = Cx+Cy
Cx = Ax+Bx = -6.36 + 0 = -6.36
Cy = Ay+By = 6.36 + 15 = 21.36

Vector C = (-6.36)² + (21.36)²; basic Pythagorean Theorem
Vector C = 22.29

So to calculate average acceleration over 8 seconds, I would think that I need to find ∆V/∆t, which would be 22.29/8, which equals 2.79 seconds, which is woefully incorrect. I could also see 6.36/8 sec to find the accel in the x direction and then figure out the accel in the y direction, but my 21.36 number is way off. I'm not understanding the neg/pos thing after reading what y'all wrote.

I'm close though, and still battling ......

6. Sep 11, 2010

Got it! Thanks for the help!!

7. Sep 11, 2010

stevebac

This part is incorrect. Vector C is the change in velocity, which means
C = B - A

Therefore,
Vector C = Cx(i)+Cy(j)
Cx = Bx - Ax= 0 - (-6.36) = 6.36
Cy = By - Ax = 15 - 6.36 = 8.64

Hence the magnitude of the velocity vector, which is the speed, is about 10.7

Edit: glad to see that you've solved it.

Last edited: Sep 11, 2010