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Ch 02: Kinematics in One Dimension

Chapter 2, Problem 2

FIGURE EX2.32 shows the acceleration graph for a particle that starts from rest at t = 0 s. What is the particle's velocity at t = 6 s?

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Hey, everyone in this problem, the acceleration plotted against time for an object is shown in the graph below. We're asked to find the change in velocity of the object from T equals zero seconds to T equals seconds. Now, the graph we're given shows the acceleration in meters per second squared on the Y axis or the vertical axis and it has the time in seconds on the horizontal axis. It shows that the acceleration starts at zero m per second squared at a time of zero seconds and increases linearly up to 12 m per second squared at 20 seconds. We're given four answer choices. Option a 120 m per second. Option B 110 m/s. Option C 135 m/s and option D 186 m/s. Now, we're given this graph of acceleration versus time and we want to find the change in velocity. So what's the connection between the velocity and an acceleration time curve? We'll recall that the change in velocity delta V is equal to the area under the V T curve. Oops not the V T curve or the A T curve, the acceleration time curve. OK. So the area under our acceleration curve on our acceleration versus time graph is gonna give us a change in velocity, which is what we're looking for. So in our problem, the change in velocity and we're looking at the change in velocity from T equals zero seconds to T equals 20 seconds. So we want to find the area under those particular points. OK? So T equals zero seconds. You know the acceleration of zero, it increases linearly to an acceleration of 12 m per second at T equals 20 seconds. We're gonna take the area under this curve and that is going to be a change in velocity between those two time points in this area. This shape is a triangle. And so the area of this triangle is gonna be one half multiplied by the base multiplied by the height. We have one half the base of our triangle where we go from zero seconds all the way up to 20 seconds on the X axis. And so the base is going to be 20 seconds multiplied by the height. OK? And the height we go from zero m per second squared all the way up to 12 m per second squared. And so the height is gonna be 12 m per second squared. If we're looking at our units, we have seconds multiplied by meters per second squared. This is gonna give us a unit of meter per second, which is what we want for velocity. We have one half multiplied by 20, multiplied by 12. We get 120 m/s and that is a change in velocity we were looking for. So we were given an A T curve, we found the area under the curve between two particular time points which gives us change in velocity between those time points. We found that the correct answer is 120 m per second, which corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.