| The position of a particle as a function of time t, is given by \[x(t)=at\,+b{{t}^{2}}-c{{t}^{3}}\] where a, b and c are constants. |
| When the particle attains zero acceleration, then its velocity will be: [JEE Main 9-4-2019 Afternoon] |
A) \[a+\frac{{{b}^{2}}}{4c}\]
B) \[a+\frac{{{b}^{2}}}{c}\]
C) \[a+\frac{{{b}^{2}}}{2c}\]
D) \[a+\frac{{{b}^{2}}}{3c}\]
Correct Answer: D
Solution :
| [d] \[x=at+b{{t}^{2}}-c{{t}^{3}}\] |
| \[\text{v}=\frac{dx}{dt}=a+2bt-3c{{t}^{2}}\] |
| \[a=\frac{dv}{dt}=2b-6ct=0\Rightarrow t=\frac{b}{3c}\] |
| \[{{^{\text{v}}}_{\left( at\,t=\frac{b}{3c} \right)}}=a+2b\left( \frac{b}{3c} \right)-3c\left( \frac{b}{3c} \right)\]\[=a+\frac{{{b}^{2}}}{3c}.\] |
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