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A particle moves in a straight line with an initial velocity of 30.0 meters per second.
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It accelerates for 5.00 seconds at a constant 30.0 meters per second squared in the direction of its initial velocity.
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What is the magnitude of the particle’s displacement after the acceleration?
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What is the magnitude of the particle’s velocity after the acceleration?
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In this two-part problem, we want to solve for the particle’s displacement after it undergoes its acceleration, which we can call 𝑑.
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And we also wanna solve for its velocity after this acceleration occurs.
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We can call this 𝑣 sub 𝑓.
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In terms of the particle’s motion, we’re told it starts out with an initial velocity of 30.0 meters per second.
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And then, for 5.00 seconds, it accelerates at 30.0 meters per second squared.
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We can come to a better understanding of the particle’s motion by drawing a graph of its velocity versus time.
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Here we have a chart of the particle’s velocity versus time, where time, in units of seconds, is on the horizontal axis and velocity, in units of meters per second, is on the vertical axis.
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We’re told that our particle begins with a velocity, we’ve called it 𝑣 sub 𝑖, of 30.0 meters per second.
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So we mark that value in as our initial velocity on our chart.
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We’re told further that our particle that accelerates at an acceleration of 30.0 meters per second squared for a time, we’ve called 𝑡 sub 𝑎, of 5.00 seconds.
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When we write out our acceleration of 30.0 meters per second squared, we understand that that means 30.0 meters per second every second.
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In other words, for every one second of time passed, we add 30.0 meters per second to our velocity.
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So if the next tick mark on our vertical axis is 60 meters per second, then we achieve that velocity at a time of one second.
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And under this continuing acceleration of 30.0 meters per second, we continue to add 30.0 meters per second to our velocity every second.
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By connecting these dots, we now have our velocity versus time curve for the particle.
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And we see that under these conditions, the maximum velocity we achieve is 180 meters per second.
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That’s our result for 𝑣 sub 𝑓, the final velocity achieved.
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In our case, we’ve solved for this result graphically.
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That’s another way of finding the same result.
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Now, we want to solve for the total displacement of the particle.
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We’ve called it 𝑑.
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We can recall that an object’s displacement is equal to the integral of its velocity with respect to time.
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And our graph is a velocity versus time curve.
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This means that if we solve for the area under our velocity versus time curve, then that total area result will give us our displacement 𝑑.
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To make calculating this total area easier, we can divide the area up into two pieces, a rectangular piece as well as a triangular piece above that.
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We can now write that our displacement 𝑑 is equal to the area of the rectangle plus the area of the triangle in units of meters.
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We can write that the area of our rectangle is equal to its height, 30.0 meters per second, times it’s width, 5.00 seconds.
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Notice the units of seconds cancel, and we’re left with units of meters.
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Then we add that area to the area of our triangle which is one-half its base, 5.00 seconds, times its height, 180 minus 30, or one 150 meters per second.
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Again, the units of seconds cancels out, and we’re left with units of meters.
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When we add these two values together, we find a result of 525 meters.
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That’s the particle’s total displacement after these five seconds of acceleration and starting at our initial velocity 𝑣 sub 𝑖.