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

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

Questions 1 - 10

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?

$49.2:\text{mph}$

$60.2:\text{mph}$

$29.2:\text{mph}$

$48.2:\text{mph}$

$39.2:\text{mph}$

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 have a $90^{\circ}$ angle between them.

For our calculations, let , and .

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

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?

$11.2:\text{mph}$

$15.2:\text{mph}$

$12.2:\text{mph}$

$13.2:\text{mph}$

$14.2:\text{mph}$

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 have a $90^{\circ}$ angle between them.

For our calculations, let , and .

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

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?

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

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

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

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

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

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.

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

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?

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

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

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

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

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

Explanation

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

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

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?

$5:\text{mph}$

$25:\text{mph}$

$10:\text{mph}$

$15:\text{mph}$

$1:\text{mph}$

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.

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

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?

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

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

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

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

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

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 have a $90^{\circ}$ angle between them.

For our calculations, let , and .

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

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?

$67.1:\uptext{mph}$

Not possible to find

$77.1:\uptext{mph}$

$30.1:\uptext{mph}$

$90:\uptext{mph}$

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 have a $90^{\circ}$ angle between them.

For our calculations, let , and .