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

Answer: (B)

8 m/s

Answer 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: $8 \, \text{m/s}$