What was the hiker's average velocity during part B of the hike? In Part A, a hiker travels four kilometers east from nine to nine-forty-five a. m. In Part B, a hiker travels six kilometers south from nine-forty-five to ten-forty-five a. m. In Part C, a hiker travels four kilometers west from ten-forty-five to eleven-fifteen a. m. In Part D, a hiker travels six kilometers north from eleven-fifteen a. m. to twelve p. m.

Question

Pedro Golden

Chemistry

7 April, 05:53

What was the hiker's average velocity during part B of the hike? In Part A, a hiker travels four kilometers east from nine to nine-forty-five a. m. In Part B, a hiker travels six kilometers south from nine-forty-five to ten-forty-five a. m. In Part C, a hiker travels four kilometers west from ten-forty-five to eleven-fifteen a. m. In Part D, a hiker travels six kilometers north from eleven-fifteen a. m. to twelve p. m.

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Good · Answer 1

You have this in Chemistry. It should be in physics, but never mind that.

A

d = 4 km due east. See note for part B.

t = 45 minutes = 3/4 hour

v = ?

d = v * t

4 = v * (3/4)

v = 4 / (3/4) = 4 * 4/3 = 16/3 = 5 1/3

So his rate of travel is 5 1/3 km/hr due east. If you do not give the direction when it says velocity, you are wrong. Velocity has direction.

B

10 45 to 11 15 is 30 minutes. 30/60 = 1/2 hour.

d = v * t

d = 6 km due south (this is actually a displacement which also have direction).

t = 1/2 hour

6 = v * (1/2)

6 / (1/2) = v

6 / (0.5) = v

v = 12 km / hr due south

Even running that's an impossible speed.