Question

A particle is moving in a straight line. The variation of position $ x $ as a function of time $ t $ is given as: 
$ x = t^3 - 6t^2 + 20t + 15 $. 
The velocity of the body when its acceleration becomes zero is:

• A. \( 6 \) m/s
• B. \( 10 \) m/s
• C. \( 8 \) m/s
• D. \( 4 \) m/s

Hint:

To find when a particle reaches maximum or minimum velocity, set the acceleration \( a = 0 \) and solve for \( t \), then substitute \( t \) in the velocity equation.

Answer 1

The Correct Option is C

Step 1: {Find velocity}
Velocity is the first derivative of displacement: \[ v = \frac{dx}{dt} = 3t^2 - 12t + 20 \] Step 2: {Find acceleration}
Acceleration is the derivative of velocity: \[ a = \frac{dv}{dt} = 6t - 12 \] Step 3: {Set acceleration to zero}
\[ 6t - 12 = 0 \] Solving for \( t \): \[ t = 2 \,s \] Step 4: {Find velocity at \( t = 2 \)}
\[ v = 3(2)^2 - 12(2) + 20 \] \[ = 12 - 24 + 20 = 8 \,m/s \] Step 5: {Verify the options}
The correct answer is (C) \( 8 \) m/s.

Answer 2

Given:
Position function of a particle: \[ x(t) = t^3 - 6t^2 + 20t + 15 \]
We are to find the **velocity** when the **acceleration becomes zero**.

Step 1: Find velocity as the first derivative of position
\[ v(t) = \frac{dx}{dt} = 3t^2 - 12t + 20 \]

Step 2: Find acceleration as the derivative of velocity
\[ a(t) = \frac{dv}{dt} = 6t - 12 \]

Step 3: Set acceleration to zero to find the time
\[ 6t - 12 = 0 \Rightarrow t = 2 \, \text{seconds} \]

Step 4: Find velocity at \( t = 2 \)
\[ v(2) = 3(2)^2 - 12(2) + 20 = 12 - 24 + 20 = 8 \, \text{m/s} \]

Final Answer:
The velocity when acceleration becomes zero is \( \boxed{8} \, \text{m/s} \).