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Common Core: High School - Number and Quantity : Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

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All Common Core: High School - Number and Quantity Resources

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Example Questions

Example Question #1 : 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?

Possible Answers:

$90\:\uptext{mph}$

Not possible to find

$67.1\:\uptext{mph}$

$30.1\:\uptext{mph}$

$77.1\:\uptext{mph}$

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 .

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

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Example Question #2 : 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?

Possible Answers:

$39.2\:\text{mph}$

$49.2\:\text{mph}$

$48.2\:\text{mph}$

$60.2\:\text{mph}$

$29.2\:\text{mph}$

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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Example Question #3 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Jack slides down a hill at $45\:\text{mph}$ , and throws a rock behind him at $8\:\text{mph}$ . How fast is the rock going?

Possible Answers:

$4\:\text{mph}$

$8\:\text{mph}$

$53\:\text{mph}$

$45\:\text{mph}$

$37\:\text{mph}$

Correct answer:

$37\:\text{mph}$

Explanation:

Since the rock is going in the opposite direction of Jack, we simply subtract the speed of the rock from how fast Jack is going down the hill.

$\text{Rock Speed}=45-8=37$

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Example Question #3 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

If an airplane is flying south at $500\:\frac{\text{km}}{\text{hr}}$ , and there are winds coming from the west at $150\:\frac{\text{km}}{\text{hr}}$ , how fast is the plane going?

Possible Answers:

$1012\:\frac{\text{km}}{\text{hr}}$

$424\:\frac{\text{km}}{\text{hr}}$

$500\:\frac{\text{km}}{\text{hr}}$

$522\:\frac{\text{km}}{\text{hr}}$

$620\:\frac{\text{km}}{\text{hr}}$

Correct answer:

$522\:\frac{\text{km}}{\text{hr}}$

Explanation:

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

Screen shot 2016 03 17 at 4.39.13 pm

Since the airplane is traveling south, and the wind is coming from the 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 the airplane 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=500 , and B=150 .

$h=\sqrt{500^2+150^2}=\sqrt{250000+22500}=\sqrt{272500}\approx522\:\frac{\text{km}}{\text{hr}}$

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Example Question #2 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Jill slides down a hill at $27\:\text{mph}$ , and throws a coin forward at $8\:\text{mph}$ . How fast is the coin going?

Possible Answers:

$39\:\text{mph}$

$35\:\text{mph}$

$15\:\text{mph}$

$19\:\text{mph}$

$25\:\text{mph}$

Correct answer:

$35\:\text{mph}$

Explanation:

Since the coin is going in the same direction as Jill, we simply add the speed of the coin and how fast Jill is going down the hill.

$\text{Coin Speed}=27+8=35$

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Example Question #4 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Bob slides down a hill at $15\:\text{mph}$ , and throws his wallet behind him at $10\:\text{mph}$ . How fast is his wallet going?

Possible Answers:

$25\:\text{mph}$

$15\:\text{mph}$

$1\:\text{mph}$

$10\:\text{mph}$

$5\:\text{mph}$

Correct answer:

$5\:\text{mph}$

Explanation:

Since Bob's wallet is going in the opposite direction, we simply subtract the speed of the wallet from how fast Bob is going down the hill.

$\text{Wallet Speed}=15-10=5$

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Example Question #6 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

John is cruising north down a river at $5\: \uptext{mph}$ , and the river has $10\: \uptext{mph}$ current due east. What is John's actual speed?

Possible Answers:

$11.2\:\text{mph}$

$15.2\:\text{mph}$

$12.2\:\text{mph}$

$14.2\:\text{mph}$

$13.2\:\text{mph}$

Correct answer:

$11.2\:\text{mph}$

Explanation:

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

Screen shot 2016 03 18 at 10.44.23 am

Since John 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 John 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=5 , and B=15 .

$h=\sqrt{5^2+10^2}=\sqrt{25+100}=\sqrt{125}\approx11.2\:\uptext{mph}$

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Example Question #3 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Jack slides down a hill at $37\:\text{mph}$ , and throws a rock ahead of him at $20\:\text{mph}$ . How fast is the rock going?

Possible Answers:

$47\:\text{mph}$

$37\:\text{mph}$

$20\:\text{mph}$

$57\:\text{mph}$

$17\:\text{mph}$

Correct answer:

$57\:\text{mph}$

Explanation:

Since the rock is going in the same direction as Jack, we simply add the speed of the rock to how fast Jack is going down the hill.

$\text{Rock Speed}=37+20=57$

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Example Question #8 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

If an airplane is flying south at $500\:\frac{\text{km}}{\text{hr}}$ , and there are winds going north at $150\:\frac{\text{km}}{\text{hr}}$ , how fast is the plane going?

Possible Answers:

$400\:\frac{\text{km}}{\text{hr}}$

$650\:\frac{\text{km}}{\text{hr}}$

$500\:\frac{\text{km}}{\text{hr}}$

$150\:\frac{\text{km}}{\text{hr}}$

$350\:\frac{\text{km}}{\text{hr}}$

Correct answer:

$350\:\frac{\text{km}}{\text{hr}}$

Explanation:

Since the airplane and the wind are going in opposite directions, we simply subtract the speed of the wind from the speed of the plane.

$\text{Airplane Speed}=500-150=350$

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Example Question #9 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

If an airplane is flying south at $1000\:\frac{\text{km}}{\text{hr}}$ , and there are winds going south at $500\:\frac{\text{km}}{\text{hr}}$ , how fast is the plane going?

Possible Answers:

$1500\:\frac{\text{km}}{\text{hr}}$

$1250\:\frac{\text{km}}{\text{hr}}$

$500\:\frac{\text{km}}{\text{hr}}$

$1000\:\frac{\text{km}}{\text{hr}}$

$250\:\frac{\text{km}}{\text{hr}}$

Correct answer:

$1500\:\frac{\text{km}}{\text{hr}}$

Explanation:

Since the airplane and the winds are going the same direction, we simply add the airplane and wind speeds together.

$\text{AIrplane Speed}=1000+500=1500$

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All Common Core: High School - Number and Quantity Resources

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Common Core: High School - Number and Quantity : Vector Representation of Quantities (Velocity, etc.): CCSS.Math.Content.HSN-VM.A.3

Study concepts, example questions & explanations for Common Core: High School - Number and Quantity

All Common Core: High School - Number and Quantity Resources

Example Questions

Example Question #1 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Example Question #2 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Example Question #3 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Example Question #3 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Example Question #2 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Example Question #4 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Example Question #6 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Example Question #3 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Example Question #8 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

Example Question #9 : Vector Representation Of Quantities (Velocity, Etc.): Ccss.Math.Content.Hsn Vm.A.3

All Common Core: High School - Number and Quantity Resources

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