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A bus makes a trip according to the position-time graph shown in the illustration. What is the average acceleration (in $\mathrm{km} / \mathrm{h}^{2} )$ of the bus for the entire 3.5 -h period shown in the graph?
$-8.3 k m / h^{2}$
Physics 101 Mechanics
Chapter 2
Kinematics in One Dimension
Motion Along a Straight Line
Alyssa J.
October 2, 2021
-8.6 km/h^2
Cornell University
University of Michigan - Ann Arbor
Simon Fraser University
University of Winnipeg
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this problem were given a position Time graph of a bus in its motion were asked to find the average acceleration of the entire problem. So if you just use the definition of the average acceleration, it's whatever you're starting speed in your final speed are two difference between those divided by the entire time for the trip. So the in this, um, graph, they give us three sections, but all we really care about is section A and section C because Section A will give us our the initial and section C will give us our final velocity. So in the data for Section A, it looks like we have a initial position zero in a final position of about 24 kilometers and then the time interval is zero to one hour. So the velocity for that section velocity for section A would be 24 minus zero, divided by the time interval one minus zero. So that gives us 24 kilometers per hour for Section C. We have an initial position of about 33 kilometers in a final position of about 27. The time interval for the starting point for Section C is about 2.2 hours and the final time is 3.5. So the velocity for section B B 27 kilometers minus the 33 divided by a difference of 3.5 hours and 10.2. So we have negative six kilometers over 1.3 hours and that gives us a velocity of negative 4.6 kilometers per hour. Your numbers might be slightly different because you're sort of estimating when you look at a graph without fine precision. So now we need to find the overall change in velocity. So we started in the first section with a velocity of 24 and we ended with a velocity of negative 4.6. So our acceleration is gonna be the final velocity minus the initial over the time interval. So negative 4.6 minus 24 divided by the total time for the whole trip, which is 3.5 hours. And that gives us an acceleration of negative 8.2 kilometers per hour squared. So again, this could very based on what you estimated for the values for both position and for time. And that would be something close to a negative a point to
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