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 18

A sprinter accelerates from rest to 9.00 m/s in 1.38 s. What is her acceleration in

  1. m/s2\textrm{m/s}^2;
  2. km/h2\textrm{km/h}^2?
  1. 6.52 m/s26.52\textrm{ m/s}^2
  2. 8.45×104 km/h28.45\times10^4\textrm{ km/h}^2
Giancoli 7th Edition, Chapter 2, Problem 18 solution video poster

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This is Giancoli Answers with Mr. Dychko. The sprinter's acceleration is the change in their velocity divided by time so that's final velocity of 9 meters per second minus initial velocity of zero. So we have 9 divided by 1.38 seconds which is 6.52 meters per second squared. In part (b), we can convert that into kilometers per hour squared by multiplying this meters per second squared by 3600 seconds for every hour all squared that gives us seconds squared here canceling with this second squared here and then we have hours squared on the bottom and then we multiply by 1 kilometer for every 1000 meters giving us 8.45 times 10 to the 4 kilometers per hour squared. So every hour, the speed of the sprinter would change by 8.45 times 10 to the 4 kilometers per hour.

By schokoladenkuchen07 on Tue, 6/29/2021 - 4:35 PM

Hello, why do we have to square the 3600?

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