A) \[b,\,-4d\]
B) \[-b,\,2c\]
C) \[b,\,2c\]
D) \[2c,\,-4d\]
Correct Answer: C
Solution :
\[y=a+bt+c{{t}^{2}}-d{{t}^{4}}\] \[\therefore \ v=\frac{dy}{dt}=b+2ct-4d{{t}^{3}}\] and \[a=\frac{dv}{dt}=2c-12d{{t}^{2}}\] Hence, at t = 0, vinitial = b and ainitial = 2c.
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