Question #256be

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Question #256be

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Answer 1 · 2015-11-14T04:30:26

$8.70$ $8.70$ units with an angle of ${16.70}^{\circ}$ $16.70^\circ$

Explanation:

The first vector is heading along the $x$ $x$ axis, with a magnitude of 4 units.

The second vector has an angle of $30^{\circ}$ $30^{\circ}$ , with a magnitude of 5 units. Decomposing this vector into its $x$ $x$ and $y$ $y$ components we have:

$x$ $x$ component: $5 cos (30) = 4.33$ $5cos(30)=4.33$
$y$ $y$ component: $5 sin (30) = 2.50$ $5sin(30)=2.50$

Let's call the first vector $\to u$ $\vec{u}$ and the second $\to t$ $\vec{t}$ .

$\to u = 4 ˆ i$ $\vec{u}=4\hat{i}$
$\to s = 4.33 ˆ i + 2.50 ˆ j$ $\vec{s}=4.33\hat{i}+2.50\hat{j}$

The sum of the two vectors will be:

$\to t = \to u + \to s = 4 ˆ i + 4.33 ˆ i + 2.50 ˆ j = 8.33 ˆ i + 2.50 ˆ j$ $\vec{t}=\vec{u}+\vec{s}=4\hat{i}+4.33\hat{i}+2.50\hat{j}=8.33\hat{i}+2.50\hat{j}$

The magnitude $∣ ∣ ∣ \to t ∣ ∣ ∣ = \sqrt{{8.33}^{2} + {2.50}^{2}} = 75.64 = 8.70$ $|\vec{t}|=\sqrt{8.33^2+2.50^2}=75.64=8.70$
The angle is $arctan (\frac{2.5}{8.33}) = {16.70}^{\circ}$ $\arctan(2.5/8.33)=16.70^\circ$

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Question #256be

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