JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2019
- Hard
- MCQ
The position of a particle as a function of time \(t\), is given by \(x\left( t \right) = at+ b{t^2} - c{t^3}\) where \(a, b\) and \(c\) are constants. When the particle attains zero acceleration, then its velocity will be
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(a + \frac{{{b^2}}}{{3c}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
X = at + b{t^2} - c{t^3}\\
V = \frac{{dx}}{{dt}} = a + 2bt - 3c{t^2}\\
a = \frac{{dv}}{{dt}} = 2b - 6ct\\
Put\,acceleration\, = 0\\
\Rightarrow \,\,t = \frac{b}{{3c}}\\
Find\,V\,at\,t = \frac{b}{{3c}}\\
V = \,a + \frac{{{b^2}}}{{3c}}
\end{array}\)
X = at + b{t^2} - c{t^3}\\
V = \frac{{dx}}{{dt}} = a + 2bt - 3c{t^2}\\
a = \frac{{dv}}{{dt}} = 2b - 6ct\\
Put\,acceleration\, = 0\\
\Rightarrow \,\,t = \frac{b}{{3c}}\\
Find\,V\,at\,t = \frac{b}{{3c}}\\
V = \,a + \frac{{{b^2}}}{{3c}}
\end{array}\)
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