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

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All Common Core: High School - Number and Quantity Resources

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Example Questions

Example Question #121 : High School: Number And Quantity

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

Possible Answers:

\displaystyle 77.1\:\uptext{mph}

Not possible to find

\displaystyle 90\:\uptext{mph}

\displaystyle 30.1\:\uptext{mph}

\displaystyle 67.1\:\uptext{mph}

Correct answer:

\displaystyle 67.1\:\uptext{mph}

Explanation:

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

Bob1Bob2

Since bob is traveling south, and the current is traveling west, they are perpendicular to each other. This means that they are \displaystyle 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 \displaystyle h=\sqrt{A^2+B^2}, where \displaystyle h is the hypotenuse, and \displaystyle A\displaystyle B are the legs of the triangle, and \displaystyle A,B have a \displaystyle 90^{\circ} angle between them.

For our calculations, let \displaystyle A=30, and \displaystyle B=60

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

 

 

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

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

Possible Answers:

\displaystyle 39.2\:\text{mph}

\displaystyle 49.2\:\text{mph}

\displaystyle 48.2\:\text{mph}

\displaystyle 60.2\:\text{mph}

\displaystyle 29.2\:\text{mph}

Correct answer:

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

For our calculations, let \displaystyle A=45, and \displaystyle B=20

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

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

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

Possible Answers:

\displaystyle 4\:\text{mph}

\displaystyle 8\:\text{mph}

\displaystyle 53\:\text{mph}

\displaystyle 45\:\text{mph}

\displaystyle 37\:\text{mph}

Correct answer:

\displaystyle 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. 

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

For our calculations, let \displaystyle A=500, and \displaystyle B=150

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

Example Question #122 : High School: Number And Quantity

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

Possible Answers:

\displaystyle 25\:\text{mph} 

\displaystyle 35\:\text{mph}

\displaystyle 15\:\text{mph}

\displaystyle 39\:\text{mph}

\displaystyle 19\:\text{mph}

Correct answer:

\displaystyle 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. 

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

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

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

Possible Answers:

\displaystyle 25\:\text{mph}

\displaystyle 5\:\text{mph}

\displaystyle 15\:\text{mph}

\displaystyle 10\:\text{mph}

\displaystyle 1\:\text{mph}

Correct answer:

\displaystyle 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. 

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

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

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

Possible Answers:

\displaystyle 11.2\:\text{mph}

\displaystyle 15.2\:\text{mph}

\displaystyle 12.2\:\text{mph}

\displaystyle 14.2\:\text{mph}

\displaystyle 13.2\:\text{mph}

Correct answer:

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

For our calculations, let \displaystyle A=5, and \displaystyle B=15

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

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

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

Possible Answers:

\displaystyle 47\:\text{mph}

\displaystyle 37\:\text{mph}

\displaystyle 20\:\text{mph}

\displaystyle 57\:\text{mph}

\displaystyle 17\:\text{mph}

Correct answer:

\displaystyle 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. 

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

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

\displaystyle 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.

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

Example Question #121 : High School: Number And Quantity

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

Possible Answers:

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

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

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

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

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

Correct answer:

\displaystyle 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.

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

All Common Core: High School - Number and Quantity Resources

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