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All Common Core: High School - Number and Quantity Resources

6 Diagnostic Tests 49 Practice Tests Question of the Day Flashcards Learn by Concept

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Flashcards: Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

Bob is cruising south down a river at $60\: \uptext{mph}$ , and the river has $30\: \uptext{mph}$ current due west. What is Bob's actual speed?

Not possible to find

$30.1\:\uptext{mph}$

$90\:\uptext{mph}$

$77.1\:\uptext{mph}$

$67.1\:\uptext{mph}$

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Flashcards: Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

Bob is cruising south down a river at $60\: \uptext{mph}$ , and the river has $30\: \uptext{mph}$ current due west. What is Bob's actual speed?

Your answer:

Not possible to find

Your answer:

$30.1\:\uptext{mph}$

Your answer:

$90\:\uptext{mph}$

Your answer:

$77.1\:\uptext{mph}$

Correct answer:

$67.1\:\uptext{mph}$

Explanation:

In order to figure this out, we need to create a picture.

Bob1 Bob2

Since bob is traveling south, and the current is traveling west, they are perpendicular to each other. This means that they are $90^{\circ}$ to each other. The next step is to use the Pythagorean Theorem in order to solve for what speed Bob is actually going.

Recall that the Pythagorean Theorem is $h=\sqrt{A^2+B^2}$ , where is the hypotenuse, and , are the legs of the triangle, and A,B have a $90^{\circ}$ angle between them.

For our calculations, let A=30 , and B=60 .

$h=\sqrt{30^2+60^2}=\sqrt{900+3600}=\sqrt{4500}\approx67.1\:\uptext{mph}$

Flashcards: Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

Amanda is cruising north down a river at $45\: \uptext{mph}$ , and the river has $20\: \uptext{mph}$ current due east. What is Amanda's actual speed?

$29.2\:\text{mph}$

$39.2\:\text{mph}$

$60.2\:\text{mph}$

$48.2\:\text{mph}$

$49.2\:\text{mph}$

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Flashcards: Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

Amanda is cruising north down a river at $45\: \uptext{mph}$ , and the river has $20\: \uptext{mph}$ current due east. What is Amanda's actual speed?

Your answer:

$29.2\:\text{mph}$

Your answer:

$39.2\:\text{mph}$

Your answer:

$60.2\:\text{mph}$

Your answer:

$48.2\:\text{mph}$

Correct answer:

$49.2\:\text{mph}$

Explanation:

In order to figure this out, we need to create a picture.

Screen shot 2016 03 17 at 4.15.01 pm

Since Amanda is traveling north, and the current is traveling east, they are perpendicular to each other. This means that they are $90^{\circ}$ to each other. The next step is to use the Pythagorean Theorem in order to solve for what speed Amanda is actually going.

Recall that the Pythagorean Theorem is $h=\sqrt{A^2+B^2}$ , where is the hypotenuse, and , are the legs of the triangle, and A,B have a $90^{\circ}$ angle between them.

For our calculations, let A=45 , and B=20 .

$h=\sqrt{45^2+20^2}=\sqrt{2025+400}=\sqrt{2425}\approx49.2\:\uptext{mph}$

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All Common Core: High School - Number and Quantity Resources

Flashcards: Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

Flashcards: Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

Flashcards: Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

Flashcards: Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

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