Giancoli's Physics: Principles with Applications, 7th Edition

2

Describing Motion: Kinematics in One Dimension

Change chapter2-1 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 28

Q

A

$23\textrm{ m/s}$

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. In this question, it will be important to get your negative signs correct in order to avoid taking the square root of a negative number which will just make your calculater explode. So we are gonna have distance is actually displacement is what we should properly be saying and it's positive 65 meters will define the car's initial direction to be positive so the car's initially going, say to the right and that will be positive 65 meters in which case it's acceleration is in the opposite direction since it's slowing down and it's traveling to the right and if it's slowing down that means the acceleration is in the opposite direction to the velocity which is to the left and so that gives an acceleration of negative 4. So this formula here is the one that contains all of the known's and the single unknown, the *v i*, the final velocity we know is zero and initial velocity is what we want to find and we'll subtract *2ad* from both sides and switch the sides around and we have *v i* squared equals negative *2ad* and so *v i* will be the square root of this, take the square root of both sides and so that means it's the square root of negative 2 times the negative 4 meters per second squared times 65—I didn't bother writing the negative signs in the calculator there because I know negative times a negative makes a positive— anyway, we end up with 23 meters per second when you have two significant figures.

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