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Finding Average Velocity

The position ...

Question

The position $x$ of a particle w.r.t time $t$ alon x-axis is given by $x = 9 t^{2} - t^{3}$ where $x$ is in metre and in second. Find What will be the position of this particle when it achieves max speed along +x direction?

74 m

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64 m

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54 m

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59 m

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Solution

The correct option is C $54 m$
The speed $v$ will be given as $v = \frac{d x}{d t} = 9 \times 2 t - 3 t^{2} = 18 t - 3 t^{2}$
this speed will be $m a x i m u m$ when $\frac{d v}{d t} = 0$ or $18 - 6 t = 0$ or $t = 3 s e c$
putting $t = 3$ in the $e x p r e s s i o n$ for $x$ we get $x = 9 \times 3^{2} - 3^{3} = 81 - 27 = 54 m e t e r$

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The position x of a particle w.r.t time t alon x-axis is given by x=9t2−t3 where x is in metre and in second. Find What will be the position of this particle when it achieves max speed along +x direction?

The correct option is C 54mThe speed v will be given as v=dxdt=9×2t−3t2=18t−3t2this speed will be maximum when dvdt=0 or 18−6t=0 or t=3secputting t=3 in the expression for x we get x=9×32−33=81−27=54meter

The position $x$ of a particle w.r.t time $t$ alon x-axis is given by $x = 9 t^{2} - t^{3}$ where $x$ is in metre and in second. Find What will be the position of this particle when it achieves max speed along +x direction?