Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
2
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: Falling Objects [neglect air resistance]
2-8: Graphical Analysis

Problem 17
A
a)7.41m/s27.41 m/s^2
b)9.60×104km/h29.60 \times 10^4 km/h^2
Giancoli 6th Edition, Chapter 2, Problem 17 solution video poster
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VIDEO TRANSCRIPT

The change in speed of the sprinter is ten meters per second as he starts from rest and he accelerates for one point three five seconds, which gives seven point four zero seven which in three significant figures is seven point four one meters per second squared, that’s the answer for part a. For part b they want us to express this in kilometers per hour squared. Let’s do it this way, let’s multiply this answer seven point four zero seven meters per second squared by one kilometer for every one thousand meters, the meters cancel leaving us with kilometers on top, times three thousand six hundred seconds for every hour and we’ll have to multiply it by this twice to get rid of both the seconds because it’s seconds squared in the bottom, the two seconds cancel with the seconds squared leaving us with hours squared on the bottom, so our answer is nine point six zero times ten raised to power four kilometers per hour squared. So if this sprinter is able to accelerate for an entire hour they’d end up going ninety six thousand kilometers and hour by the end of the hour.

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