Mathematics, 04.06.2020 13:09 alinegonzalez0027

A car travels 10km southeast and then 15 km in a direction 60 degrees north of east. What is the magnitude of the car's resultant vector?

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Answer from: saren47

The magnitude of the car's resultant vector is 15.72.

Given that,

A car travels 10km southeast,

And then 15 km in a direction 60 degrees north of east.

We have to find,

The magnitude of the car's resultant vector.

According to the question,

To resultant direction calculate by using the sum of vectors following all the steps given below.

A car travels 10 km southeast,

And then 15 km in a direction 60 degrees north of east.

The first vector has module is 10 and angle is 315°,

In the south direction, the angle is 270° and east is 360°,

So the angle in southeast,

$\rm = \dfrac{270+360}{2}\\\\= \dfrac{630}{2}\\\\= 315 \ degree$

The second vector has module 15 and angle = 40°

Decompose both vectors in their horizontal and vertical component:

The horizontal component of the first vector is,

$\rm = 10 \times cos(315) = 10 \times 0.70= 7.07$

The vertical component of the first vector is,

$= \rm 10 \times sin(315) = -7.07$

The horizontal component of the second vector is,

$\rm = 15 \times cos(60) = 15 \times 0.5= 7.5$

The vertical component of the second vector is,

$\rm = 15 \times sin(40) = 15 \times 0.86= 12.99$

The sum of the horizontal component of the resultant vector is,

= 7.07+7.5=14.57

And the sum of the vertical component of the resultant vector is,

= -7.07+12.99=5.91

The magnitude of the vector (14.57km, 5.91km) is:

Therefore,

The magnitude of the car's resultant vector is,

$\rm Resulatant = \sqrt{(14.57)^2+ (5.91)^2}\\\\ Resulatant = \sqrt{212.28+ 34.92} \\\ \\ Resulatant = \sqrt{247.20}\\\\ Resulatant = 15.72$

Hence, The required magnitude of the car's resultant vector is 15.72.

For more details refer to the link given below.

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Answer from: DESIRE44030

15.85km

Step-by-step explanation:

The car starts at the point (0,0)

Let's define east as the positive x-axis and north as the positive y-axis.

first, the car travels 10km southeast, so the angle is -45° from the east.

Then the new position of the car is:

(10km*cos(-45°), 10km*sin(-45°)) = (7.07km, -7.07km)

Now, from this point the car travels 15km with an angle of 60° (counting from the east)

So the new position is:

(7.07km + 15km*cos(60°), -7.07km + 15km*sin(60°)) = (14.57km , 5.91km)

The magnitude of a vector (x, y) is √(x^2 + y^2)

So the magnitude of the vector (14.57km , 5.91km) is:

√((14.57km)^2 + (5.91km)^2) = 15.85km

Answer from: Quest

Will show energy on x axis time on the y axis feel free to ask any questions

Answer from: Quest

answer: correct

step-by-step explanation:

What is the slope of the line in the graph? slope is 1 answer=1

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A car travels 10km southeast and then 15 km in a direction 60 degrees north of east. What is the mag...