A) \[{{b}_{{}^\circ }}\]
B) \[{{b}_{1}}\]
C) \[{{b}_{2}}\]
D) \[2{{b}_{2}}\]
Correct Answer: D
Solution :
[d] \[x={{b}_{0}}+{{b}_{1}}t+{{b}_{2}}{{t}^{2}}\] \[\frac{dx}{dt}=v={{b}_{1}}+2{{b}_{2}}t\] \[a=\frac{dv}{dt}=2{{b}_{2}}\]Note: \[v=\frac{dx}{dt}\] \[a=\frac{dv}{dt}\] |
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