A) \[b,-4d\]
B) \[-b,2c\]
C) \[b,2c\]
D) \[2c,-4d\]
Correct Answer: C
Solution :
\[y=a+bt+c{{t}^{2}}-d{{t}^{4}}\] \[\therefore \]\[v=\frac{dv}{dt}=b+2ct-4d{{t}^{3}}\] and \[a=\frac{dv}{dt}=2c-12\,d{{t}^{2}}\] Hence, at t = 0, \[{{v}_{initial}}=b\]and \[{{a}_{initial}}=2c.\]You need to login to perform this action.
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