91Ó°ÊÓ

Consider the following vector,

$B=−4\widehat{i}+4\widehat{j}$The levelled diagram for the given vector is as follows,

The magnitude of the vector is calculated as follows,

$B=\sqrt{{\left(B_x\right)}^2+{\left(B_y\right)}^2}$Substitute $−4$for $B_x$and 4 for$B_y$.

$B=\sqrt{{(−4)}^2+{\left(4\right)}^2}=4\sqrt{2}$The angle is calculated as follows,

$\mathrm{θ}={\tan }^{−1}\frac{\left|B_y\right|}{\left|B_x\right|}$Substitute 4 for $\left|B_x\right|$and 4 for $\left|B_y\right|$.

$\mathrm{θ}={\tan }^{−1}\frac{4}{4}={45}^{°}$

Therefore, the magnitude of the resultant is $2.2\text{cm}$and the direction is ${27}^{°}$with the $x$axis.

Consider the following vector,

$\stackrel{→}{r}=(−2.0\widehat{i}−1.0\widehat{j})\text{cm}$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 $−2\text{cm}$for $r_x$and $−1\text{cm}$for $r_y$.

$r=\sqrt{{(−2\text{cm})}^2+{(−1\text{cm})}^2}=\sqrt{5\text{cm}}=2.2\text{cm}$The angle is calculated as follows,

$\mathrm{θ}={\tan }^{−1}\frac{\left|r_y\right|}{\left|r_x\right|}$Substitute $2\text{cm}$for $\left|r_x\right|$and $1\text{cm}$for $\left|r_y\right|$.

$\mathrm{θ}={\tan }^{−1}\frac{1\text{cm}}{2\text{cm}}={27}^{°}$

Therefore, the magnitude of the resultant is $2.2\text{cm}$and the direction is ${27}^{°}$with the $x$axis.

$100.5\text{m}/\text{s}$

Consider the following vector,

$\stackrel{→}{v}=(−10\widehat{i}−100\widehat{j})\text{m}/\text{s}$The leveled 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 $−10\text{m}/\text{s}$for $v_x$and $−100\text{m}/\text{s}$for $v_y$.

$v=\sqrt{{(−10\text{m}/\text{s})}^2+{(−100\text{m}/\text{s})}^2}=\sqrt{10100{\text{m}}^2/{\text{s}}^2}=100.5\text{m}/\text{s}$The angle is calculated as follows,

$\mathrm{θ}={\tan }^{−1}\frac{\left|v_y\right|}{\left|v_x\right|}$Substitute $10\text{m}/\text{s}$for $\left|v_x\right|$and $100\text{m}/\text{s}$for $\left|v_y\right|$.

$\mathrm{θ}={\tan }^{−1}\frac{100\text{m}/\text{s}}{10\text{m}/\text{s}}={84}^{°}$

Therefore, the magnitude of the resultant is $100.5\text{m}/\text{s}$and the direction is ${84}^{°}$with the $x$axis.

Consider the following vector,

$a=(20\widehat{i}+10\widehat{j})\text{m}/{\text{s}}^2$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 $20\text{m}/{\text{s}}^2$for $a_x$and $10\text{m}/{\text{s}}^2$for$a_y$.

$a=\sqrt{{\left(20\text{m}/{\text{s}}^2\right)}^2+{\left(10\text{m}/{\text{s}}^2\right)}^2}=22\text{m}/{\text{s}}^2$The angle is calculated as follows,

$\mathrm{θ}={\tan }^{−1}\frac{\left|a_y\right|}{\left|a_x\right|}$Substitute $20\text{m}/{\text{s}}^2$for $\left|a_x\right|$and $10\text{m}/{\text{s}}^2$for $\left|a_y\right|$.

$\mathrm{θ}={\tan }^{−1}\frac{10\text{m}/{\text{s}}^2}{20\text{m}/{\text{s}}^2}={27}^{°}$

Therefore, the magnitude of the resultant is $22\text{m}/{\text{s}}^2$and the direction is ${27}^{°}$with the $x$axis.