Let
acceleration of the particle = a
time to travel from A to B = t
speed at point A = VA
speed at point C = VC
According to constant acceleration motion formulas
"|AB| = V_A\\cdot t+a\\frac{t^2}{2}""a=\\frac{V_C-V_A}{t+10}""60=2t+a\\frac{t^2}{2}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ (1)""a=\\frac{5-2}{t+10}=\\frac{3}{t+10}\\ \\ \\ (2)" Substituting a from (2) into equation (1):
"60=2t+\\frac{3t^2}{2(t+10)}""120(t+10)=4t(t+10)+3t^2""120t+1200=4t^2+40t+3t^2""7t^2-80t-1200=0" using quadratic formula:
"t_{1,2}=\\frac{-(-40)\\pm \\sqrt{40^2-7\\cdot(-1200)}}{7}=""\\frac{40\\pm \\sqrt{10000}}{7}=\\frac{40\\pm 100}{7}"
"t_1,=20,\\ t_2=-60\/7" Since t > 0 t = t1 = 20s
From equation (2)
"a=\\frac{3}{20+10}=\\frac{3}{30}=\\frac{1}{10}m\/s^2" Answer:
"\\frac{1}{10}m\/s^2"
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