A particle moves in 2D motion such that its position and time related to each other as - - t2 - 10 - - - - - 1- Find velocity of particle at - - 2 sec in m-s

Given: x=t^2+10⇒ v_x=dxdt=2t so, v_x|_@ t=2 sec=2×2=4 Also, y=t 1⇒ v_y=dydt=1 So, at t=2 sec, v_y=1 Thus, velocity of particle is given by v=v_xî+v_yĵ |v|=√(v_ ...

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Byju's Answer

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 $x = t^{2} + 10, y = t - 1$ . Find velocity of particle at $t = 2 sec (in m/s)$

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Solution

Given:
$x = t^{2} + 10 \Rightarrow v_{x} = \frac{d x}{d t} = 2 t$
so, $v_{x} |_{@ t = 2 sec} = 2 \times 2 = 4$
Also, $y = t - 1 \Rightarrow v_{y} = \frac{d y}{d t} = 1$
So, at $t = 2 sec$ , $v_{y} = 1$
Thus, velocity of particle is given by
$\to v = v_{x}^i + v_{y}^j$
$| \to v | = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{4^{2} + 1^{2}} = \sqrt{17}$
Hence, option $(C)$ is correct.

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A particle moves in 2D motion such that its position and time related to each other as x=t2+10,y=t−1. Find velocity of particle at t=2 sec (in m/s)

Given: x=t2+10⇒ vx=dxdt=2t so, vx|@ t=2 sec=2×2=4 Also, y=t−1⇒ vy=dydt=1 So, at t=2 sec, vy=1 Thus, velocity of particle is given by →v=vx^i+vy^j |→v|=√v2x+v2y=√42+12=√17 Hence, option (C) is correct.

A particle moves in $2 D$ motion such that its position and time related to each other as $x = t^{2} + 10, y = t - 1$ . Find velocity of particle at $t = 2 sec (in m/s)$