Question

The distance covered by a particle moving in a straight line path at time $ t $ (in seconds) is given by $ s = (t^3 - 6t^2 + 3t + 4) \, \text{m}$. The velocity of the particle when its acceleration is zero, will be

• A. 3 m/s
• B. 42 m/s
• C. -12 m/s
• D. -9 m/s

Hint:

To find velocity and acceleration in motion problems, use the derivatives of the position function with respect to time.

Answer 1

The Correct Option is A

Step 1: Find the expression for velocity.
Velocity \( v \) is the first derivative of distance \( s \) with respect to time \( t \): \[ v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 3t + 4) \] \[ v = 3t^2 - 12t + 3 \]
Step 2: Find the expression for acceleration.
Acceleration \( a \) is the derivative of velocity \( v \) with respect to time: \[ a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 3) \] \[ a = 6t - 12 \]
Step 3: Find when acceleration is zero.
Set \( a = 0 \): \[ 6t - 12 = 0 \] \[ t = 2 \, \text{seconds} \]
Step 4: Find the velocity when acceleration is zero.
Substitute \( t = 2 \) into the velocity equation: \[ v = 3(2)^2 - 12(2) + 3 = 12 - 24 + 3 = -9 \, \text{m/s} \]
Conclusion:
The velocity of the particle when its acceleration is zero is \( -9 \, \text{m/s} \).
Conclusion:
The correct answer is (D) -9 m/s.