The position x of a particle with respect to time t along x -axis is given by x = 9t2-t3 where x is in metres and t in seconds. What will be the position of this particle when it achieves maximum speed along the +x direction?

Question

The position x of a particle with respect to time t along x -axis is given by x = 9t2-t3 where x is in metres and t in seconds. What will be the position of this particle when it achieves maximum speed along the +x direction? (A) 54 m (B) 81 m (C) 24 m (D) 32 m

Tardigrade Admin · Accepted Answer

Given: x=9 t2-t3 ...(i) Speed v=(d x/d t)=(d/d t)(9 t2-t3)=18 t-3 t2 For maximum speed, (d v/d t)=0 ⇒ 18-6 t=0 ∴ t=3 s ∴ x max =81 m -27 m =54 m (From x=9 t2-t3 ).

Q. The position $x$ of a particle with respect to time t along $x$ -axis is given by $x = 9 t^{2} - t^{3}$ where $x$ is in metres and $t$ in seconds. What will be the position of this particle when it achieves maximum speed along the $+ x$ direction?

54 m

81 m

24 m

32 m

Given : $x = 9 t^{2} - t^{3}$ ...(i)
Speed $v = \frac{d x}{d t} = \frac{d}{d t} (9 t^{2} - t^{3}) = 18 t - 3 t^{2}$
For maximum speed, $\frac{d v}{d t} = 0$
$\Rightarrow 18 - 6 t = 0$
$∴ t = 3 s$
$∴ x_{m a x} = 81 m - 27 m = 54 m$
(From $x = 9 t^{2} - t^{3}$ ).

Q. The position x of a particle with respect to time t along x -axis is given by x=9t2−t3 where x is in metres and t in seconds. What will be the position of this particle when it achieves maximum speed along the +x direction?