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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 $27 \mathrm{m} / \mathrm{s},$ and the average southward velocity has a magnitude of 17 $\mathrm{m} / \mathrm{s}$ . What is the average velocity (magnitude and direction) for the entire trip?
$v_{a v g .}=16 \mathrm{m} \cdot \mathrm{s}^{-1}$
Physics 101 Mechanics
Chapter 2
Kinematics in One Dimension
Motion Along a Straight Line
University of Washington
Simon Fraser University
University of Sheffield
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Okay. So in this problem we have a car that is travelling north with a velocity of 27 meters per second and it travels north for a while before it turns around and goes south at 17 meters per second. Which since south is the opposite direction, is going to be negative 17 meters per second. Now, in terms of what we know about the time, the only thing we know here is that this 27 meters per second was happening for 3/4 of the road trip. And this, uh, 17 meters per second south was only happening for 1/4 of the road trip. So in order to find the average velocity, I'm actually going to break this trip up average. I'm going to break this trip up, um, into four different pieces. So since the North Velocity was taking place for 3/4 of the trip, that 27 meters per second will get recorded three times, some adding the other my 1st 3 speeds, then that 17 meters per second is the last fourth. And to find the average, I divide by the quantity of numbers. And if I add all of those together and divide by four. We end up with positive 16 meters per second, meaning the average velocity of the car for the strip is 16 meters per second to the north.
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