JEE Mains · Physics · STD 11 - 2. motion in straight line
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
- A \(a + \frac{{{b^2}}}{{4c}}\)
- B \(a + \frac{{{b^2}}}{{c}}\)
- C \(a + \frac{{{b^2}}}{{2c}}\)
- D \(a + \frac{{{b^2}}}{{3c}}\)
Answer & Solution
Correct Answer
(D) \(a + \frac{{{b^2}}}{{3c}}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} X = at + b{t^2} - c{t^3}\\ V = \frac{{dx}}{{dt}} = a + 2bt - 3c{t^2}\\ a = \frac{{dv}}{{dt}} = 2b - 6ct\\ Put\,acceleration\, = 0\\ \Rightarrow \,\,t = \frac{b}{{3c}}\\ Find\,V\,at\,t = \frac{b}{{3c}}\\ V = \,a + \frac{{{b^2}}}{{3c}} \end{array}\)
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