91Ó°ÊÓ

The formula to find the magnitude of resultant of two vectors $\stackrel{→}{A}$and$\stackrel{→}{B}$
is as follows,

$R=\sqrt{{\left(A\right)}^2+{\left(B\right)}^2}$ Here $R$is the magnitude of the resultant vector.

Consider the following vector,

$\stackrel{→}{A}=3\widehat{i}+7\widehat{j}$The levelled diagram for the given vector is as follows,

The magnitude of the vector is calculated as follows,

$A=\sqrt{{\left(A_x\right)}^2+{\left(A_y\right)}^2}$Substitute 7 for $A_x$and 3 for $A_y$.

$A=\sqrt{{\left(7\right)}^2+{\left(3\right)}^2}=\sqrt{58}=7.6$The angle is calculated as follows,

$\mathrm{θ}={\tan }^{−1}\frac{\left|A_y\right|}{\left|A_x\right|}$Substitute 3 for $\left|A_x\right|$and 7 for $\left|A_y\right|$.

$\mathrm{θ}={\tan }^{−1}\frac{7}{3}={66.8}^{°}={67}^{°}$

Therefore, the magnitude of the resultant is $7.6$ and the direction is ${67}^{°}$with the $x$axis.

Consider the following vector,

$a=(−2\widehat{i}+4.5\widehat{j})\text{m}/{\text{s}}^2$The levelled diagram for the given vector is as follows,

The magnitude of the vector is calculated as follows,

$a=\sqrt{{\left(a_x\right)}^2+{\left(a_y\right)}^2}$Substitute $−2\text{m}/{\text{s}}^2$for $a_x$and $4.5\text{m}/{\text{s}}^2$for $a_y$.

$a=\sqrt{{\left(−2\text{m}/{\text{s}}^2\right)}^2+{\left(4.5\text{m}/{\text{s}}^2\right)}^2}=4.9\text{m}/{\text{s}}^2$The angle is calculated as follows,

$\mathrm{θ}={\tan }^{−1}\frac{\left|a_y\right|}{\left|a_x\right|}$Substitute $2\text{m}/{\text{s}}^2$for $\left|a_x\right|$and $4.5\text{m}/{\text{s}}^2$for $\left|a_y\right|$.

$\mathrm{θ}={\tan }^{−1}\frac{4.5\text{m}/{\text{s}}^2}{2\text{m}/{\text{s}}^2}={66}^{°}$Therefore, the magnitude of the resultant is $4.9\text{m}/{\text{s}}^2$and the direction is ${66}^{°}$with the $x$axis.

Consider the following vector,

$\stackrel{→}{v}=(14\widehat{i}−11\widehat{j})\text{m}/\text{s}$The levelled diagram for the given vector is as follows,

The magnitude of the vector is calculated as follows,

$v=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}$Substitute $14\text{m}/\text{s}$for $v_x$and $−11\text{m}/\text{s}$for $v_y$.

$v=\sqrt{{(14\text{m}/\text{s})}^2+{(−11\text{m}/\text{s})}^2}=18\text{m}/\text{s}$The angle is calculated as follows,

$\mathrm{θ}={\tan }^{−1}\frac{\left|v_y\right|}{\left|v_x\right|}$Substitute $14\text{m}/\text{s}$for $\left|v_x\right|$and $11\text{m}/\text{s}$for $\left|v_y\right|$.

$\mathrm{θ}={\tan }^{−1}\frac{11\text{m}/\text{s}}{14\text{m}/\text{s}}={38}^{°}$

Therefore, the magnitude of the resultant is $18\text{m}/\text{s}$and the direction is ${38}^{°}$with the $x$axis.

Consider the following vector,

$\stackrel{→}{r}=(−2.2\widehat{i}−3.3\widehat{j})\text{m}$The levelled diagram for the given vector is as follows,

The magnitude of the vector is calculated as follows,

$r=\sqrt{{\left(r_x\right)}^2+{\left(r_y\right)}^2}$Substitute $−2.2\text{m}$for $r_x$and $−3.3\text{m}$for $r_y$.

$r=\sqrt{{(−2.2\text{m})}^2+{(−3.3\text{m})}^2}=4\text{m}$The angle made with the y axis is calculated as follows,

$\mathrm{θ}={\tan }^{−1}\frac{\left|r_y\right|}{\left|r_x\right|}$Substitute $2.2\text{m}$for $\left|r_x\right|$and $3.3\text{m}$for $\left|r_y\right|$.

$\mathrm{θ}={\tan }^{−1}\frac{3.3\text{m}}{2.2\text{m}}={56}^{°}$Therefore, the angle made with the axis is calculated as follows,

${90}^{°}−{56}^{°}={34}^{°}$Therefore, the magnitude of the resultant is $4\text{m}$and the direction is ${34}^{°}$with the$x$ axis.