Average Velocity of a Car on a North-South Trip - Simple Velocity Question

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A car travels north for three-fourths of the time at 27 m/s and south for one-fourth of the time at 17 m/s. The average velocity is calculated using the formula v_avg = (3/4 * 27 + 1/4 * -17), resulting in +16 m/s. The discussion clarifies that the average velocity should be reported as a magnitude and direction rather than a signed value. The final answer confirms the calculations are correct, emphasizing the importance of presenting the result appropriately.
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Homework Statement



A car makes a trip due north for three-fourths of the time and due south one-fourth of the time. The average northward velocity has a magnitude of 27 m/s, and the average southward velocity has a magnitude of 17 m/s. What is the average velocity (magnitude and direction) for the entire trip?

Homework Equations



##\displaystyle v_{avg} = \frac{x}{t}##

The Attempt at a Solution


[/B]
##\displaystyle v_{avg} = \frac{x}{t} = \frac{v_1t_1 + v_2t_2}{t} = \frac{v_1(\frac{3}{4}t) + v_2(\frac{1}{4}t)}{t} = \frac{3}{4}v_1 + \frac{1}{4}v_2 = \frac{3}{4}27 + \frac{1}{4}(-17) = +16~m/s##

Is this the correct answer for the average velocity over the whole trip? If not, what am I doing wrong?
 
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Is the southward velocity 7 or 17?
 
jbriggs444 said:
Is the southward velocity 7 or 17?
Oops, it's 17 m/s. I fixed it.
 
In that case, your work looks correct.

Edit: The question asks for magnitude and direction rather than a signed result.
 
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jbriggs444 said:
In that case, your work looks correct.

Edit: The question asks for magnitude and direction rather than a signed result.

Alright, I'll take that into account. Thanks!
 
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