Question

2)A velocity vector has components 36 m/s westward and 22 m/s northward. What are the magnitude and direction of this vector?

Answer 1

The magnitude of the velocity vector is $\boxed{42.19\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}$ and the direction of the vector is $\boxed{58.57^\circ }$ north of west.

Further Explanation:

Given:  

The component of velocity vector in the westward direction is $22\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}$.  

The component of velocity vector in the north direction is $36\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}$.

Concept:  

Since the velocity components are in the west direction and the north direction, they are perpendicular to each other. The magnitude of the actual velocity vector can be determined as resultant of the two components that are perpendicular to each other.  

The resultant of the vector can be expressed as:

$R = \sqrt {v_1^2 + v_2^2 +2{v_1}{v_2}\cos\theta }$

Here, ${v_1}\,\&\,{v_2}$ are the components of velocity vector and $\theta$ is the angle between them.  

Substitute the values in the above expression  

$\begin{aligned}R&= \sqrt {{{\left( {36} \right)}^2} + {{\left( {22}\right)}^2} + 2 \times 36\times 22 \times\cos 90^\circ }\\&=\sqrt {1296 + 484}\\&= \sqrt {1780} \,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 42.19\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The direction of the velocity vector is given by:

$\begin{aligned}\alpha&={\tan ^{ - 1}}\left( {\frac{{{v_2}}}{{{v_1}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{36}}{{22}}} \right)\\&= 58.57^\circ \,{\text{north}}\,{\text{of}}\,{\text{west}}\\\end{aligned}$

Thus, the magnitude of the velocity vector is $\boxed{42.19\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}$  and the direction of the vector is $\boxed{58.57^\circ }$ north of west.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Vector and Scalar

Keywords:  Velocity vector, component, westward, northward, magnitude of velocity, direction of the velocity, resultant, perpendicular, two components of velocity.

Answer 2

The magnitude:
 →
| a | = √ ( 22² + 36² ) = √ ( 484 + 1296 ) = √ 1780 = 42.19 m/s
The direction:
tan α = 22/36 = 0.61111
α = tan^(-1) 0.61111 = 31.43 ° = 31° 26`
Answer:
The magnitude is 42.19 m/s and the direction is 31° 26`north from west.