A rhinoceros is at the origin of coordinates at t1=0. For the next 12 seonds, the rhino's average velocity has x component -3.6 m/s and y component 3.9 m/s. Assuming no acceleration, at time t2=12 seconds, what are the x and y coordinates of the rhino? How far is the rhino from the origin?
I did V = sq rt (-3.6^2 + 3.9^2) to get the resultant, then multiplied that by time to get displacement. Is that right?
And how can I find the coordinates? Thanks very much!!!!
2007-09-02 12:42:35 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Given that (x1,y1) = (0,0)
x2 coordinate = Vx * T
x2 = (-3.6 m/s) * 12 s
x2 = -43.2 m
y2 coordinate = Vy * T
y2 = (3.9 m/s) * 12s
y2 = 46.8 m
(x,y)2 = (-43.2 m, 46.8 m)
D = sq rt (-43.2^2 + 46.8^2)
D = 63.7 m
Which should be the same value you got the way you described
2007-09-02 15:03:01 · answer #1 · answered by skeptik 7 · 0⤊ 0⤋
seek the attitude that belows to the gap of two m perspective ? = 2 / (2r?) * one hundred eighty = 2 / (12?) * one hundred eighty = 9.5493° x = r cos(?) = 6 * cos(9.5493°) = 5.ninety two m y = r sin(?) = 6 * sin(9.5493°) = 0.ninety 9 m hint: circumference of the circle = 2*r * ? = 12 * ? you may only verify the answer x² + y² = 6²
2016-12-31 10:22:20 · answer #2 · answered by ? 3 · 0⤊ 0⤋
on graph paper.
2007-09-02 12:50:38 · answer #3 · answered by Anonymous · 0⤊ 1⤋