A man travels 30 m along the direction of 3 | vedclass

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(D) The initial displacement vector $\vec{S}_1$ is along the direction of $3 \hat{i} + 4 \hat{j}$. The unit vector is $\hat{u} = \frac{3 \hat{i} + 4 \

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) The initial displacement vector $\vec{S}_1$ is along the direction of $3 \hat{i} + 4 \hat{j}$. The unit vector is $\hat{u} = \frac{3 \hat{i} + 4 \hat{j}}{\sqrt{3^2 + 4^2}} = \frac{3 \hat{i} + 4 \hat{j}}{5}$.Thus,$\vec{S}_1 = 30 	imes \left(\frac{3 \hat{i} + 4 \hat{j}}{5}ight) = 18 \hat{i} + 24 \hat{j} \ m$.Let the second displacement be $\vec{S}_2 = d \hat{n}$,where $\hat{n}$ is a unit vector perpendicular to $\vec{S}_1$. Since $\vec{S}_1 = 18 \hat{i} + 24 \hat{j}$,a perpendicular vector is proportional to $4 \hat{i} - 3 \hat{j}$.So,$\vec{S}_2 = k(4 \hat{i} - 3 \hat{j})$. The magnitude is $d = |k| \sqrt{4^2 + (-3)^2} = 5|k|$.The total displacement $\vec{S}_{net} = \vec{S}_1 + \vec{S}_2 = (18 + 4k) \hat{i} + (24 - 3k) \hat{j}$.Since the net displacement is along the $x$-axis,the $y$-component must be zero:$24 - 3k = 0 \implies k = 8$.Then $\vec{S}_2 = 8(4 \hat{i} - 3 \hat{j}) = 32 \hat{i} - 24 \hat{j}$.The magnitude $d = |\vec{S}_2| = \sqrt{32^2 + (-24)^2} = \sqrt{1024 + 576} = \sqrt{1600} = 40 \ m$.

$A$ man travels $30 \ m$ along the direction of $3 \hat{i} + 4 \hat{j}$ and then moves '$d$' meters perpendicular to the initial direction such that his total displacement is along the $x$-axis. What is the value of '$d$' in meters?

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