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 these
Correct Answer: B
Solution :
\[x=4(t-2)+a{{(t-2)}^{2}}\] At \[t=0,\,x=-8+4a=4a-8\] \[v=\frac{dx}{dt}=4+2a(t-2)\] At \[t=0,\ \ v=4-4a=4(1-a)\] But acceleration,\[a=\frac{{{d}^{2}}x}{d{{t}^{2}}}=2a\]You need to login to perform this action.
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