Question:
A particle moves according to the law \( s = t^3 - 6t^2 + 9t + 25 \). The displacement of the particle at the time when its acceleration is zero is
A particle moves according to the law \( s = t^3 - 6t^2 + 9t + 25 \). The displacement of the particle at the time when its acceleration is zero is
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When acceleration is zero, always first find time using \( a = 0 \), then substitute it into the displacement equation.
Updated On: Mar 28, 2026
- \( 0 \) units
- \( -27 \) units
- \( 27 \) units
- \( 9 \) units
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The Correct Option is C
Solution and Explanation
Step 1: Find velocity.
\[ v = \frac{ds}{dt} = 3t^2 - 12t + 9 \] Step 2: Find acceleration.
\[ a = \frac{dv}{dt} = 6t - 12 \] Step 3: Set acceleration equal to zero.
\[ 6t - 12 = 0 \Rightarrow t = 2 \] Step 4: Find displacement at \( t = 2 \).
\[ s = (2)^3 - 6(2)^2 + 9(2) + 25 \] \[ = 8 - 24 + 18 + 25 = 27 \] Step 5: Conclusion.
The displacement of the particle is \( 27 \) units.
\[ v = \frac{ds}{dt} = 3t^2 - 12t + 9 \] Step 2: Find acceleration.
\[ a = \frac{dv}{dt} = 6t - 12 \] Step 3: Set acceleration equal to zero.
\[ 6t - 12 = 0 \Rightarrow t = 2 \] Step 4: Find displacement at \( t = 2 \).
\[ s = (2)^3 - 6(2)^2 + 9(2) + 25 \] \[ = 8 - 24 + 18 + 25 = 27 \] Step 5: Conclusion.
The displacement of the particle is \( 27 \) units.
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