A particle moves in 2D motion such that its position and time related to each other as - - 4t2 - 4t-3 - - - 2- - 1- Find velocity of particle at - - 1 sec in m-s

Given: x=4t^2+4t+3⇒ v_x=dxdt=8t+4 so, v_x|_@ t=1 sec=8×1+4=12 Also, y=2t 1⇒ v_y=dydt=2 So, at t=1 sec, v_y=2 Thus, velocity of particle is given by v=v_xî+v_yĵ ...

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Standard XII

Physics

Instantaneous Velocity

A particle mo...

Question

A particle moves in $2 D$ motion such that its position and time related to each other as $x = 4 t^{2} + 4 t + 3, y = 2 t - 1$ . Find velocity of particle at $t = 1 sec (in m/s)$

\sqrt{158}

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\sqrt{140}

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\sqrt{148}

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\sqrt{144}

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Solution

The correct option is C $\sqrt{148}$
Given:
$x = 4 t^{2} + 4 t + 3 \Rightarrow v_{x} = \frac{d x}{d t} = 8 t + 4$
so, $v_{x} |_{@ t = 1 sec} = 8 \times 1 + 4 = 12$
Also, $y = 2 t - 1 \Rightarrow v_{y} = \frac{d y}{d t} = 2$
So, at $t = 1 sec$ , $v_{y} = 2$
Thus, velocity of particle is given by
$\to v = v_{x}^i + v_{y}^j$
$| \to v | = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{12^{2} + 2^{2}} = \sqrt{148}$
Hence, option $(B)$ is correct.

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A particle moves in 2D motion such that its position and time related to each other as x=4t2+4t+3,y=2t−1. Find velocity of particle at t=1 sec (in m/s)

The correct option is C √148Given: x=4t2+4t+3⇒ vx=dxdt=8t+4 so, vx|@ t=1 sec=8×1+4=12 Also, y=2t−1⇒ vy=dydt=2 So, at t=1 sec, vy=2 Thus, velocity of particle is given by →v=vx^i+vy^j |→v|=√v2x+v2y=√122+22=√148 Hence, option (B) is correct.

A particle moves in $2 D$ motion such that its position and time related to each other as $x = 4 t^{2} + 4 t + 3, y = 2 t - 1$ . Find velocity of particle at $t = 1 sec (in m/s)$