First slide

Rectilinear Motion

Question

The relation $\sqrt{3 x} = 3 t - 6$ describes the position of a particle in one direction where ‘x’ is in meters and t in sec . The displacement when velocity is zero, is……….

Easy

A

-12m
B

5m
C

zero
D

24m

Solution

Given that $\sqrt{3 x} = 3 t - 6.......... (1)$
$\Rightarrow 3 x = 9 t^{2} + 36 - 36 t$
Differentiate on both sides with respect to time
$\begin{array}{l} ∴ 3 (\frac{d x}{d t}) = 18 t + 0 - 36 \\ \Rightarrow 3 V = 18 t - 36 \\ If V = 0 then 18 t - 36 = 0 \\ \Rightarrow 18 t = 36 \\ \Rightarrow t = 2 S \dots \dots \dots \dots . . (2) \end{array} a s v = 6 t - 12 a c c e l e r a t i o n o f t h e p a r t i c l e i s a = \frac{d v}{d t} = 6 m s^{- 2} = c o n s \tan t a t t = o v e l o c i t y o f t h e p a r t i c l e i s v = - 12 m s^{- 1} d i s p l a c e m e n t o f t h e b o d y s = u t + \frac{a t^{2}}{2} b y t h e t i m e t = 2 s e c v e l o c i t y b e c o m e s z e r o s = - 12 \times 2 + \frac{6 \times 4}{2} = - 12 m$
$∴$ from equation …….(1)
At $t = 0$ the initial position of the particle $x_{1} = 12 m$
At $t = 2 S$ the final position of the partial $x_{2} = 0$
$∴$ Displacement S=change in position
$= x_{2} - x_{1} = - 12 m$

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