A) \[b+2ct,2c\]
B) \[b,2c\]
C) \[2c,b\]
D) \[b+2c,2c\]
Correct Answer: B
Solution :
Here, \[S=a+bt+c{{t}^{2}}\] ?...(i) Differentiate equation (i) with respect to we, get. \[V=\frac{ds}{dt}=\frac{d}{dt}(a)+b\frac{d}{dt}(t)+\frac{c.d}{dt}({{t}^{2}})\] \[\frac{ds}{dt}=b+2ct.\] \[\left( \because \,\,\frac{d}{dt}+{{t}^{n}}=n{{t}^{n-1}} \right)\] \[\therefore \] Initial Velocity \[={{\left| \frac{ds}{dt} \right|}_{t=0}}=b\] Differentiate equation (ii) with respect to t, we get, \[\Rightarrow \] \[a=\frac{d\upsilon }{dt}=\frac{d}{dt}(b)+2c\frac{d}{dt}(t)\] \[\Rightarrow \] \[a=\frac{d\upsilon }{dt}=2c\] \[\Rightarrow \] Initial acceleration \[{{\left| \frac{d\upsilon }{dt} \right|}_{t=0}}=2c\]
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