Mastering Physics Solutions: Biking Vectors

Mastering Physics Solutions: Biking Vectors

On January 5, 2012, in Chapter 03: Motion in Two Dimensions, by Mastering Physics Solutions

Biking Vectors

Part A = SQRT(1.06) or 1.0296
Part B = (0.0, 1.1)
Part C = (-0.5, 0.0)
Part D = (0.0, -0.2)
Part E = (-0.5, 0.9)
Part F = 60.9°

Solutions Below:

A student bikes to school by traveling first dN = 1.10 miles north, then dW = 0.500 miles west, and finally dS = 0.200 miles south.

Part A

If a bird were to start out from the origin (where the student starts) and fly directly (in a straight line) to the school, what distance db would the bird cover?

db = SQRT(dx2 + dy2) = SQRT((-0.500)2 + (1.10 – 0.200)2)

db = SQRT(0.25 + 0.81)

db = SQRT(1.06) or 1.0296

Part B

Let the vector dN be the displacement vector corresponding to the 1st leg of the trip. Express dN in component form.

dN = (0.0, 1.1)

Part C

For the 2nd leg (dW):

dW = (-0.5, 0.0)

Part D

For the 3rd leg (dS):

dS = (0.0, -0.2)

Part E

The displacement vector for the bird, db can be written as dN + dW + dS. Express this vector in component form:

Just add the north and south directions together (1.1 – 0.2 = 0.9)

db = (-0.5, 0.9)

Part F

Find φ, the angle North of West of the path followed by the bird.

Remember SOHCAHTOA. In this case, Tan(φ) = Opposite / Adjacent. North of west means y is the opposite, x the adjacent. So:

Tan(φ) = 0.9 / 0.5
Tan(φ) = 1.8

φ = Tan-1(1.8)

φ = 60.9°

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2 Responses to Mastering Physics Solutions: Biking Vectors

  1. Tartles says:

    Thank you

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