91影视

Using the vector algebra and simple mathematical operations, the unknown quantities can be found. Here the vector addition is to be used to find the magnitude of vector.

In this problem, the displacement vector $\stackrel{鈬赌}{d}$can ne written as,

$$\begin{gathered} \stackrel{鈫�}{d}=\stackrel{鈫�}{x}\hat{\mathrm{i}}+\stackrel{鈫�}{y}\hat{\mathrm{j}}+\stackrel{鈫�}{z}\hat{\mathrm{k}} \\ \left(\mathrm{i}\right) \end{gathered}$$

With magnitude,

$$\begin{gathered} \left|\stackrel{鈫�}{d}\right|=\sqrt{x^2+y^2+z^2} \\ \left(\mathrm{ii}\right) \end{gathered}$$

Given are,

role="math" localid="1656308404966" $$\begin{gathered} x=1300\mathrm{m} \\ \mathrm{y}=2200\mathrm{m} \\ \mathrm{z}=410\mathrm{m} \end{gathered}$$

Substituting the values of x, y, and z, in equation (i) the displacement from start to landing point is given by,

$\stackrel{鈫�}{d}=\left(1300\mathrm{m}\right)\hat{\mathrm{i}}+\left(2200\mathrm{m}\right)\hat{\mathrm{j}}+\left(-410\mathrm{m}\right)\hat{\mathrm{k}}$

In this case the final position and the initial position of the man is same. Thus, the net displacement is zero. Therefore, the vector ${\hat{d}}^{\text{'}}$for the return trip which can be written as,

$\stackrel{鈫�}{d^{\text{'}}}=\left(-1300\mathrm{m}\right)\hat{\mathrm{i}}+\left(-2200\mathrm{m}\right)\hat{\mathrm{j}}$

Thus the magnitude ofis given by,

$$\begin{gathered} d^{,}=\sqrt{{\left(-1300\mathrm{m}\right)}^2+{\left(-2200\mathrm{m}\right)}^2} \\ {\stackrel{鈫�}{d}}^{\text{'}}=2.56脳{10}^3\mathrm{m}/\mathrm{s} \end{gathered}$$