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:
- 4 m/s
- 8 m/s
- 10 m/s
- 6 m/s
The Correct Option is B
Approach Solution - 1
To determine the velocity of the particle when its acceleration becomes zero, we first need to understand the relationship between position, velocity, and acceleration for this motion.
- Given the position of the particle as a function of time: \(x = t^3 - 6t^2 + 20t + 15 \, \text{m}\)
- The velocity \((v)\) is the first derivative of the position with respect to time:
\(v = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 20t + 15)\)
Calculating the derivative, we get: \(v = 3t^2 - 12t + 20\)
- The acceleration \((a)\) is the first derivative of velocity (or the second derivative of position) with respect to time:
\(a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t + 20)\)
Calculating the derivative, we get: \(a = 6t - 12\)
- We need the acceleration to be zero to find the corresponding time:
\(6t - 12 = 0\)
Solving for \(t\), we get: \(6t = 12 \quad \Rightarrow \quad t = 2\)
- Substitute \(t = 2\) into the velocity expression to find the velocity when the acceleration is zero:
\(v = 3(2)^2 - 12(2) + 20\)
\(v = 3(4) - 24 + 20\)
\(v = 12 - 24 + 20\)
\(v = 8 \, \text{m/s}\)
- Therefore, the velocity of the body when its acceleration becomes zero is 8 m/s.
Hence, the correct answer is 8 m/s.
Approach Solution -2
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: \( 8 \, \text{m/s} \)
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