Question

A particle is moving in a straight line. The variation of position \( x \) as a function of time \( t \) is given as\[x = (t^3 - 6t^2 + 20t + 15) \, \text{m}.\]The velocity of the body when its acceleration becomes zero is:

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

Answer 1

The Correct Option is B

Given the position function:  
\(x = t^3 - 6t^2 + 20t + 15\)

Step 1: Find the velocity \( v \):  
The velocity is the first derivative of the position function with respect to time:  
\(v = \frac{dx}{dt} = 3t^2 - 12t + 20\)

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

Step 3: When is the acceleration zero?  
Set the acceleration to zero to find the time at which the acceleration becomes zero:  
\(6t - 12 = 0 \implies t = 2 \, \text{sec}\)

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

Thus, the velocity of the body when its acceleration becomes zero is.  \( 8 \, \text{m/s} \)

The Correct Answer is:  \( 8 \, \text{m/s} \)