Question:
For a particle moving on a straight line, it is observed that the distance \(S\) at time \(t\) is given by
\[
S=6t-\frac{t^3}{2}
\]
The maximum velocity during the motion is
For a particle moving on a straight line, it is observed that the distance \(S\) at time \(t\) is given by
\[
S=6t-\frac{t^3}{2}
\]
The maximum velocity during the motion is
Show Hint
Velocity is the derivative of displacement with respect to time. After finding \(v=\frac{dS}{dt}\), maximize the velocity expression.
Updated On: Jun 26, 2026
- \(3\)
- \(6\)
- \(9\)
- \(12\)
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The Correct Option is B
Solution and Explanation
Step 1: Differentiate displacement to get velocity.
Given, \[ S=6t-\frac{t^3}{2} \] Velocity is \[ v=\frac{dS}{dt} \] So, \[ v=\frac{d}{dt}\left(6t-\frac{t^3}{2}\right) \] \[ v=6-\frac{3t^2}{2} \]
Step 2: Find the maximum value of velocity.
The velocity function is \[ v=6-\frac{3t^2}{2} \] Since \[ \frac{3t^2}{2}\geq 0 \] for all real values of \(t\), we get \[ 6-\frac{3t^2}{2}\leq 6 \] Thus, the maximum possible value of velocity is \[ 6 \] This occurs at \[ t=0 \]
Step 3: Final conclusion.
Hence, the maximum velocity during the motion is \[ \boxed{6} \]
Given, \[ S=6t-\frac{t^3}{2} \] Velocity is \[ v=\frac{dS}{dt} \] So, \[ v=\frac{d}{dt}\left(6t-\frac{t^3}{2}\right) \] \[ v=6-\frac{3t^2}{2} \]
Step 2: Find the maximum value of velocity.
The velocity function is \[ v=6-\frac{3t^2}{2} \] Since \[ \frac{3t^2}{2}\geq 0 \] for all real values of \(t\), we get \[ 6-\frac{3t^2}{2}\leq 6 \] Thus, the maximum possible value of velocity is \[ 6 \] This occurs at \[ t=0 \]
Step 3: Final conclusion.
Hence, the maximum velocity during the motion is \[ \boxed{6} \]
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