Problem

The speed of propagation of the action potential …

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Problem 11 Medium Difficulty

A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min. (a) What was her average speed? (b) If the straight-line distance from her home to the university is 10.3 km in a direction $25.0^{\circ}$ south of east, what was her average velocity? (c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?

Answer

(a) $\overline{v}=11.1 \mathrm{m} / \mathrm{s}$
(b) $\overline{v}=9.54 \mathrm{m} / \mathrm{s}$
(c) Speed: $\overline{v}=0.855 \mathrm{m} / \mathrm{s} \quad$ Velocity: $\overline{v}=0 \mathrm{m} / \mathrm{s}$

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Physics 101 Mechanics

College Physics for AP® Courses

Chapter 2

Kinematics

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Video Transcript

so in this problem of student is traveling from home to the university for part A. She travels 12 kilometers in 18 minutes and it asked what the average speed is In order to calculate the average speed, we have to change kilometers into meters by multiplying by 1000 and we have to change minutes into seconds by multiplying by 60. Then we used the average speed equation distance divided by time to solve for the average speed, which is 11.1 meters per second. Part be asked for the average velocity. Velocity is different from speed and that has direction. So instead of being distance over time is a change in position over time. So in this case the change in position was 10.3 kilometers, which we then change into meters. But the amount of time to do this was the same because it's the same like travel. So then we take our change in position and divide. By the time we find that her average velocity was 9.54 meters per second, never being more precise, we would say that it was in the direction of 25 degrees south of east like stated in the problem now, Part C says that she stays at school and then goes back seven and 1/2 hours after she left home. So in order to figure out what the average speed and velocity for this entire trip is, we have to first figure out the total distance. So if it took 12 kilometers to go to school and 12 kilometers to go back, the total distance is 24 kilometers, which we then change into meters. We then have to take our seven and 1/2 hours of time and change that in two seconds by multiplying by 60 times 60 which is 3600. So we find that she traveled a distance of 24,000 meters in 2700 seconds. So using distance divided by time, we confined the speed of 0.89 meters per second for velocity. We have to think of the change in position. So if she went back to where she started, her change in position was zero. She actually didn't change her position at all and so we would say that the velocity is also zero meters per second

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