**I understand this is a a lot to read through so I appreciate anyone who even t

Question

**I understand this is a a lot to read through so I appreciate anyone who even takes the time to try.

Q: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?

Relevant equations
?v = V_f - V_i (vectors)

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 don't understand how to get this at all :(. I tried several different methods, but to no avail.

Step 2: Write out V_f. Vector V_f = 15m/s
I understand this since its just a given.

Step 3: Calculate ?V. ?V= 6.36m/s xhat + 8.64m/s yhat.
My understanding is that 6.36 is positive simply because its traveling in a positive direction on the graph. I don't understand where 8.64 came from.

Step 4&5 are more straight forward and I'm sure I can figure that out if I understood the steps leading up to it.

Explanation / Answer

Vi = 9 cos 145 i + 9 sin 145 j = -6.36 i + 6.36 j This expresses the initial speed in its vector components Vf = 15 sin 90 j = 15 j a = (Vf - Vi) / t this is the definition of average acceleration where a and V are vector quantities Vf - Vi = 15 j - (-6.36 i + 6.36 j) = 6.36 i + 8.64 j It might help to draw a diagram depicting Vi and Vf and then draw a vector from the tip of Vi to the tip Vj and you should get a vector similar to 6.36 i + 8.64 j - this is just the vector diagram for "subtracting" 2 vectors - for further clarification label the vectors A, B and C and note that B - A = C is the same as B = C + A

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