MHT CET · Physics · Motion In One Dimension
The position ' \(x\) ' of a particle varies with a time as \(x=a t^2-b t^3\) where ' \(a\) ' and ' \(b\) ' are constants. The acceleration of the particle will be zero at
- A \(\frac{2a}{3b}\)
- B \(\frac{a}{b}\)
- C \(\frac{a}{3b}\)
- D zero
Answer & Solution
Correct Answer
(C) \(\frac{a}{3b}\)
Step-by-step Solution
Detailed explanation
\(x=a t^2-b t^3\)
Differentiating the displacement, we get velocity
\(\mathrm{V}=2 \mathrm{at}-3 \mathrm{bt} \mathrm{t}^2\)
Differentiating, we get acceleration
\(A=2 a-6 b t\)
Substituting \(\mathrm{A}=0\)
\(\begin{aligned}
0 & =2 \mathrm{a}-6 \mathrm{bt} & \therefore \quad 6 \mathrm{bt}=2 \mathrm{a} \\
\therefore \quad \mathrm{t} & =\frac{2 \mathrm{a}}{6 \mathrm{~b}}=\frac{\mathrm{a}}{3 \mathrm{~b}} &
\end{aligned}\)
Differentiating the displacement, we get velocity
\(\mathrm{V}=2 \mathrm{at}-3 \mathrm{bt} \mathrm{t}^2\)
Differentiating, we get acceleration
\(A=2 a-6 b t\)
Substituting \(\mathrm{A}=0\)
\(\begin{aligned}
0 & =2 \mathrm{a}-6 \mathrm{bt} & \therefore \quad 6 \mathrm{bt}=2 \mathrm{a} \\
\therefore \quad \mathrm{t} & =\frac{2 \mathrm{a}}{6 \mathrm{~b}}=\frac{\mathrm{a}}{3 \mathrm{~b}} &
\end{aligned}\)
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