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A particle is moving such that s=t^(3)-6...

A particle is moving such that `s=t^(3)-6t^(2)+18t+9`, where s is in meters and t is in meters and t is in seconds. Find the minimum velocity attained by the particle.

A

`29 ms^(-1)`

B

`5 ms^(-1)`

C

`6 ms^(-1)`

D

`12 ms^(-1)`

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The correct Answer is:
To find the minimum velocity attained by the particle given the displacement function \( s(t) = t^3 - 6t^2 + 18t + 9 \), we will follow these steps: ### Step 1: Differentiate the displacement function to find the velocity function. The velocity \( v(t) \) is the first derivative of the displacement \( s(t) \) with respect to time \( t \). \[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 18t + 9) \] ### Step 2: Calculate the derivative. Using the power rule of differentiation: \[ v(t) = 3t^2 - 12t + 18 \] ### Step 3: Set the velocity function equal to zero to find critical points. To find the minimum velocity, we need to find the critical points by setting \( v(t) = 0 \): \[ 3t^2 - 12t + 18 = 0 \] ### Step 4: Simplify the equation. Dividing the entire equation by 3 gives: \[ t^2 - 4t + 6 = 0 \] ### Step 5: Use the quadratic formula to find the roots. The quadratic formula is given by: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1, b = -4, c = 6 \): \[ t = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \] \[ t = \frac{4 \pm \sqrt{16 - 24}}{2} \] \[ t = \frac{4 \pm \sqrt{-8}}{2} \] Since the discriminant is negative (\(-8\)), there are no real roots, which indicates that the velocity does not become zero at any real time \( t \). ### Step 6: Analyze the velocity function. To find the minimum value of the velocity function, we can complete the square for the quadratic expression: \[ v(t) = 3(t^2 - 4t + 6) \] \[ = 3((t - 2)^2 + 2) \] ### Step 7: Determine the minimum velocity. The expression \((t - 2)^2\) is always non-negative, and its minimum value is 0 when \( t = 2 \). Therefore, the minimum velocity occurs at: \[ v(2) = 3(0 + 2) = 6 \text{ m/s} \] ### Conclusion: The minimum velocity attained by the particle is \( 6 \text{ m/s} \). ---

To find the minimum velocity attained by the particle given the displacement function \( s(t) = t^3 - 6t^2 + 18t + 9 \), we will follow these steps: ### Step 1: Differentiate the displacement function to find the velocity function. The velocity \( v(t) \) is the first derivative of the displacement \( s(t) \) with respect to time \( t \). \[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 18t + 9) \] ...
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DC PANDEY ENGLISH-MOTION IN A PLANE-Check point 3.5
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