Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
Describing Motion: Kinematics in One Dimension
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2-1 to 2-3: Speed and Velocity
2-4: Acceleration
2-5 and 2-6: Motion at Constant Acceleration
2-7: Freely Falling Objects
2-8: Graphical Analysis

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 21

A car moving in a straight line starts at x=0x = 0 at t=0t = 0. It passes the point x=25.0 mx = 25.0 \textrm{ m} with a speed of 11.0 m/s at t=3.00 st = 3.00 \textrm{ s}. It passes the point x=385 mx = 385 \textrm{ m} with a speed of 45.0 m/s at t=20.0 st = 20.0 \textrm{ s}. Find

  1. the average velocity, and
  2. the average acceleration, between t=3.00 st = 3.00 \textrm{ s} and t=20.0 st = 20.0 \textrm{ s}.
  1. 21.2 m/s21.2\textrm{ m/s}
  2. 2.00 m/s22.00\textrm{ m/s}^2
Giancoli 7th Edition, Chapter 2, Problem 21 solution video poster

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This is Giancoli Answers with Mr. Dychko. Average velocity is the change in position over change in time. So at t equals 20 seconds, we have a position of 385 meters and then at the time of 3 seconds, the position is 25 meters. So a difference in positions divided by differences in times gives us 21.2 meters per second is the average velocity. For the acceleration, it's the change in velocity divided by time and we have to assume that these speeds that we are given are both in the positive direction both velocities. So we have 45 meters per second at time 20 seconds minus 11 meters per second at the time of 3 seconds and difference in velocities divided by difference in time gives us a positive acceleration of 2.00 meters per second squared.

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