Question

Find a vector v whose magnitude is 4 and whose component in the i direction is twice the component in the j direction.

The answer is (8√5 i + 4√5j)/5.

Could someone please give me a detailed explanation on how to do the problem? Thanks

Answer 1

The vector v with a magnitude of 4 and i-component being twice the j-component can be found by solving for y in the equation of magnitude based on the given conditions and then determining the i and j components accordingly. The result is the vector v = (8√5 i + 4√5 j) / 5.

Step-by-step explanation:

To find a vector v whose magnitude is 4, and whose i-component is twice its j-component, we can let the i-component be 2y and the j-component be y. Using the Pythagorean theorem for two-dimensional vectors, we can write the equation for magnitude as v = √((2y)^2 + y^2). Given that the magnitude is 4, the equation becomes:

4 = √(4y^2 + y^2)

4 = √(5y^2)

16 = 5y^2

y^2 = ⅔

y = √(⅔)

y = √(⅔)

y = √(⅔)

y = √(⅔)

y = √(⅔)

y = √(⅔)

(2.0 s) gives us the direction in unit vector notation. The magnitude of the acceleration is à(2.0 s)| = √√5.0² + 4.0² + (24.0)² = 24.8 m/s².

Using the value of y we calculated, the components of vector v are:

i-component = 2y = 2(⅔) = 8√5 / 5

j-component = y = ⅔ = 4√5 / 5

So the vector v can be expressed as v = (8√5 i + 4√5 j) / 5.

Answer 2

Start with

`\vec v=x\,\vec\imath+y\,\vec\jmath`

as a template for the vector
`\vec v`. Its magnitude is 4, so

`\|\vec v\|=√(x^2+y^2)=4`

Its component in the
`\vec\imath` direction is twice the component in the
`\vec\jmath` direction, which means

`x=2y`

So we have

`√((2y)^2+y^2)=√(5y^2)=4\implies y^2=\frac{16}5\implies y=\pm\frac4{\sqrt5}`

and

`x=\pm\frac8{\sqrt5}`

Lastly, rationalize the denominator:

`\frac1{\sqrt5}\cdot(\sqrt5)/(\sqrt5)=\frac{\sqrt5}5`

So we end up with two possible answers,

`\vec v=\pm\left(\frac{8\sqrt5}5\,\vec\imath+\frac{4\sqrt5}5\,\vec\jmath\right)`