Question

The position $x$ of a particle with respect to time '$t$' along x- axis is given by $x = 9t^2 - t^3$ where $x$ is in metres and $t$ in seconds. The position of this particle when it achieves maximum speed along the x direction is

• A. $24 \text{ m}$
• B. $32 \text{ m}$
• C. $54 \text{ m}$
• D. $81 \text{ m}$

Hint:

In calculus-based kinematics, remember the hierarchy: Position $\to$ Velocity $\to$ Acceleration. To find the "maximum" of any level, set the derivative (the level below it) to zero.

Answer 1

The Correct Option is C

To find the position when the speed is maximum, we must first find the expression for velocity and then determine when its rate of change (acceleration) is zero. Step 1: Find the Velocity ($v$) Velocity is the first derivative of position with respect to time: $$v = \frac{dx}{dt} = \frac{d}{dt}(9t^2 - t^3)$$ $$v = 18t - 3t^2$$

Step 2: Find the time for Maximum Speed Speed is maximum when the acceleration ($a$) is zero. Acceleration is the derivative of velocity: $$a = \frac{dv}{dt} = \frac{d}{dt}(18t - 3t^2)$$ $$a = 18 - 6t$$ Setting $a = 0$: $$18 - 6t = 0 \implies 6t = 18 \implies t = 3 \text{ s}$$

Step 3: Calculate Position ($x$) at $t = 3 \text{ s}$ Substitute $t = 3$ back into the original position equation: $$x = 9(3)^2 - (3)^3$$ $$x = 9(9) - 27$$ $$x = 81 - 27 = 54 \text{ m}$$ Therefore, the particle is at a position of $54 \text{ m}$ when it reaches its maximum speed.