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Yuki888 [10]
3 years ago
14

Suppose you are driving due east, traveling a distance of 1500 m in 2 minutes. You then turn due north and travel the same dista

nce in the same time. What can be said about the average speeds and the average velocities for the two segments of the trip?
(a) The average speeds are the same, and the average velocities are the same.
(b) The average speeds are the same, but the average velocities are different.
(c) The average speeds are different, but the average velocities are the same.
Mathematics
1 answer:
velikii [3]3 years ago
6 0

Answer:

B

Step-by-step explanation:

There are essentially two kind of quantities, the scalar quantities and the vector quantities. While vector quantities are detailed and described according to their magnitude and directions, the scalar quantities are described in terms of their magnitude alone.

Hence we say that a scalar quantity possesses magnitude only but not direction. To adequately account for the average speed and the average velocities, we need to know the quantities that gave rise to them. While the average speed is measured by the total distance divided by the total time, the average velocity is measured by the division of the total displacement over the total time.

While average speed is a scalar quantity, average velocity is a vector quantity. Hence the speed is the same irrespective of the direction while the average velocity is different as we had a change in direction according to the question

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Answer:

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General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
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  • Addition Property of Equality
  • Subtraction Property of Equality

<u>Algebra I</u>

  • Coordinates (x, y)
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  • Exponential Rule [Rewrite]:                                                                           \displaystyle b^{-m} = \frac{1}{b^m}

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Implicit Differentiation

Basic Power Rule:

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Step-by-step explanation:

<u>Step 1: Define</u>

\displaystyle \sqrt{x} - \sqrt{y} = -1

Point (1, 4)

<u>Step 2: Differentiate</u>

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  2. [Implicit Differentiation] Basic Power Rule:                                                 \displaystyle \frac{1}{2}x^{\frac{1}{2} - 1} - \frac{1}{2}y^{\frac{1}{2} - 1}\frac{dy}{dx} = 0
  3. [Implicit Differentiation] Simplify Exponents:                                               \displaystyle \frac{1}{2}x^{\frac{-1}{2}} - \frac{1}{2}y^{\frac{-1}{2}}\frac{dy}{dx} = 0
  4. [Implicit Differentiation] Rewrite [Exponential Rule - Rewrite]:                   \displaystyle \frac{1}{2x^{\frac{1}{2}}} - \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0
  5. [Implicit Differentiation] Isolate <em>y</em> terms:                                                       \displaystyle -\frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}
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Unit: Implicit Differentiation

Book: College Calculus 10e

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