A) The initial velocity of particle is 4
B) The acceleration of particle is 2a
C) The particle is at origin at t = 0
D) None of the above
Correct Answer: B
Solution :
Rate of change of displacement is velocity. \[\left( v=\frac{dx}{dt} \right)\] Given, \[x=4(t-2)+a{{(t-2)}^{2}}\] Using\[\frac{d}{dx}{{x}^{n}}=n{{x}^{n-1}},\], we have \[v=\frac{dx}{dt}=4+2a(t-2)\] At \[t=0,\] \[v=4(1-a)\] Acceleration \[a=\frac{{{d}^{2}}x}{d{{t}^{2}}}\] \[a=\frac{{{d}^{2}}x}{d{{t}^{2}}}=2a\]
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