The position vector of a moving body at any i | vedclass

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(C) The position vector is given by $\overrightarrow{r} = 5t^2 \hat{i} - 5t \hat{j}$.The velocity vector $\overrightarrow{v}$ is the derivative of pos

Vedclass · Accepted Answer

$5 \sqrt{17} 	ext{ m/s}$,making an angle of $	an^{-1} 4$ with $-$ve $	ext{Y}$ axis

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(C) The position vector is given by $\overrightarrow{r} = 5t^2 \hat{i} - 5t \hat{j}$.The velocity vector $\overrightarrow{v}$ is the derivative of position with respect to time: $\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} = 10t \hat{i} - 5 \hat{j}$.At $t = 2 	ext{ s}$,the velocity vector is $\overrightarrow{v} = 10(2) \hat{i} - 5 \hat{j} = 20 \hat{i} - 5 \hat{j} 	ext{ m/s}$.The magnitude of velocity is $|\overrightarrow{v}| = \sqrt{(20)^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = \sqrt{25 	imes 17} = 5\sqrt{17} 	ext{ m/s}$.To find the direction,let $	heta$ be the angle with the negative $	ext{Y}$-axis. From the vector components,$	an 	heta = \frac{|v_x|}{|v_y|} = \frac{20}{5} = 4$. Thus,$	heta = 	an^{-1} 4$ with the negative $	ext{Y}$-axis.

The position vector of a moving body at any instant of time is given as $\overrightarrow{r} = (5t^2 \hat{i} - 5t \hat{j}) \text{ m}$. The magnitude and direction of velocity at $t = 2 \text{ s}$ is,

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