Can you teach me how to find the displacement and distance from a v vs t graphs area. For example like the question should have a graph that starts at -12m/s and goes to 48m/s in 50s. ---Chart Description: * Type: Line chart. * Title: velocity east vs. time. * X-axis: * Label: time (s) * Scale: Ranges from 0 to 50, with major ticks at 10, 20, 30, 40, 50. The grid lines are spaced every 1 unit. * Y-axis: * Label: velocity east (m/s) * Scale: Ranges from -20 to 60, with major ticks at -20, -10, 0, 10, 20, 30, 40, 50, 60. The grid lines are spaced every 2 units. * Data Series: A single blue straight line. * The line starts near (0, -12). * The line passes through (10, -2). * The line passes through (20, 8). * The line passes through (30, 18). * The line passes through (40, 28). * The line passes through (50, 38). * The line extends from the left edge of the graph (at time = 0) to the right edge (at time = 50). * The line represents a positive slope, indicating that the velocity east is increasing linearly with time.

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Welcome to this lesson on finding displacement and distance from velocity-time graphs. Today we'll analyze a specific example where velocity changes from negative 12 meters per second to positive 38 meters per second over 50 seconds. The key concept is that the area under a v-t graph represents displacement, while the total area regardless of sign gives us distance traveled. To analyze this graph properly, we first need to find where the velocity equals zero. This is the point where the line crosses the time axis. Looking at our velocity line, we can see it has the equation v equals t minus 12. Setting this equal to zero, we get t minus 12 equals zero, which gives us t equals 12 seconds. This crossing point at 12 seconds is crucial because it divides our motion into two phases: negative velocity before 12 seconds, and positive velocity after. Now we calculate the areas under the graph in two sections. The first area is below the time axis, from 0 to 12 seconds. This forms a triangle with base 12 seconds and height negative 12 meters per second. Using the triangle area formula, we get negative 72 meters. The second area is above the time axis, from 12 to 50 seconds. This triangle has base 38 seconds and height 38 meters per second, giving us 722 meters. Notice that the negative area represents motion in the opposite direction, while the positive area represents motion in the forward direction. To find displacement, we sum the areas with their signs. Displacement equals negative 72 meters plus 722 meters, which gives us 650 meters. This positive result means the object has moved 650 meters in the positive direction from its starting position. The negative area represents backward motion during the first 12 seconds, while the larger positive area represents forward motion for the remaining time. The net effect is a displacement of 650 meters forward. For distance, we take the absolute values of both areas and add them together. Distance equals 72 meters plus 722 meters, giving us 794 meters total. This represents the total path length traveled, regardless of direction. To summarize: displacement is 650 meters, which tells us the net change in position, while distance is 794 meters, which tells us how far the object actually traveled. Remember, displacement considers direction and can be positive or negative, while distance is always positive as it represents total path length.