This is a Pythagoras one, right? I will use the sum of the squares of 2 and 3 and the square root of the answer of those which will be length of the hypotenuse and thus the length of the vector...
I thank thee with my hat in my hand... I am just apalled that I didnt try Pythagoras... which tells me I do not fully understand this work so I will go and STUDY SOME MORE...
I Have been trying to figure out what they did to get that I even tried to use the original distances - but Iam still trying ... Maybe I am just missing something.
I got initial velocity vector 9m by calculating 9cos 45 ( as direction was NE) that gave me 6.36 as x-axis is negative for the direction given I got xhat -6.36 and yhat will then be 6.36 both in m/s.
Final velocity vectors was yhat 15m/s as x is 0 for that direction which was North.
From that...
Oh my word - I will read through that chapter again - doing this through a distance learning institution and it seems the more I try to figure this the more distance there comes between me and understanding how this works! Thank you for your time and replies
Thank you so much for replying. I will try that and then make sure I understand why I had to work with that magnitude. I am studying through a distance-education institution so I rely on Google and Youtube and this forum
Thank you for answering
I know that a change in direction brings about acceleration. I presume that is why the acceleration is not constant? I will try a different equation.
1. A car is traveling 9m/s Northwest. 8 seconds later it has rounded a corner and is now heading North at 15m/s.
This was a question from my textbook and was an example question - so they supplied answers. I was able to work through all of it finding everything except the last question.
They...