A particle moves on a line with a constant acceleration of 4ft/second. Find its equation of motion if s=2ft and v=-ft/sec, when t=2sec
Newton motion second law
s=ut+12at2s=ut+\frac{1}{2}at^2s=ut+21at2
2=4×2+12×a×222=4\times2+\frac{1}{2}\times a\times2^22=4×2+21×a×22
Solve it
2−8=12a×222-8=\frac{1}{2}a\times2^22−8=21a×22
−6×2=22a-6\times2=2^2a−6×2=22a
a=−3ft/sec2a=-3ft/sec^2a=−3ft/sec2
Where
v=4ft/secv=4ft/secv=4ft/sec
s=2ftu=4ft/seca=−3ft/sec2s=2ft\\u=4ft/sec\\a=-3ft/sec^2s=2ftu=4ft/seca=−3ft/sec2
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