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Not sure how to approach this "simple" problem (picture).?
I'm guessing I'll have to find the equations for x and y and use those to find what x(1) and y(1) are and then use those to find the magnitude and direction of the resultant vector. I'm working on the solution like this for now but I want to make sure that is the correct way to approach this problem (for step 1 at least).
1 Answer
- 7 years ago
Here are the answers I got:
(a) x = 1+2t^2, so v = 4t, v(1) = 4 m/s
y = 2+3t^2, v = 6t, v(1) = 6 m/s
theta = arctan(6/4) = 56.3 degrees
(b) average velocity for x axis is 12m/s
average velocity for y axis is 18m/s
so the overall average velocity is (12+18)/2 = 15m/s
(c) average acceleration for x axis is 4m/s^2
average acceleration for y axis is 6m/s^2
so overall average acceleration is (4+6)/2 = 5m/s^2
Am I correct?