Question

The position of a particle moving along the X-axis as a function of time t (in seconds) is given by $X = t^{3}-9t + 2$. The magnitude of the acceleration of the particle at $t=3$ seconds is:

• A. 56
• B. 36
• C. 18
• D. 20

Hint:

To find acceleration from position: Differentiate twice!

Answer 1

The Correct Option is C

Step 1: Concept
Velocity ($v$) is the first derivative of position ($dX/dt$), and acceleration ($a$) is the first derivative of velocity ($dv/dt$).

Step 2: Differentiation
$v = \frac{d}{dt}(t^3 - 9t + 2) = 3t^2 - 9$.
$a = \frac{d}{dt}(3t^2 - 9) = 6t$.

Step 3: Calculation
At $t = 3$ seconds, $a = 6(3) = 18\text{ m/s}^2$.

Step 4: Conclusion
The magnitude of acceleration is 18. Final Answer: (C)