Giancoli's Physics: Principles with Applications, 7th 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: 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 34
Q

# A space vehicle accelerates uniformly from 85 m/s at $t = 0$ to 162 m/s at $t = 10.0 \textrm{ s}$. How far did it move between $t = 2.0 \textrm{ s}$ and $t = 6.0 \textrm{ s}$?

A
$460\textrm{ m}$

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. We are going to calculate the distance traveled by the space vehicle during this time interval by finding the average velocity during this time interval between 2 and 6 seconds and multiplying by 4 seconds— the difference between 2 and 6. So the final speed will be the speed at 6 seconds and the initial speed in this formula is gonna be the speed at 2 seconds and we'll calculate each speed using this formula where v f and v i mean just in general the speed at the end of whatever time interval we are looking at and the speed at the beginning; in this case, this initial speed is 85 meters per second and this v f is something we'll have to figure out. First we need to know what the acceleration is so we know that it changes its speed from 85 meters per second initially at time 0 up to 162 meters per second at time 10 seconds and so this gives us an acceleration of 7.7 meters per second squared and then we find the speed at 6 seconds by taking the initial speed of 85 meters per second plus this acceleration times 6 seconds— 131.2 meters per second— and then the speed at 2 seconds which is 85 again plus 7.7 times 2 seconds which is a 100. 4 meters per second and then take that average add them together, divide by 2 gives us the average speed times the time interval of 4 seconds gives a distance covered of 460 meters with two significant figures.

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