Question - Estimating Train Distance with Graph
mathphysicsvelocity calculationgraph analysistrain distance estimationunderestimate and overestimate
Solution:
The image contains a graph showing the velocity of a train over time and two questions:a) Use four strips of equal width to estimate the distance the train traveled in the first 20 seconds.b) Is your answer to part a) an underestimate or an overestimate?To solve part a), we need to estimate the area under the graph from 0 to 20 seconds using the four strips. Since precise measurements of the strips are not possible from the image, I'll explain the procedure.The x-axis likely represents time in seconds, and the y-axis represents velocity in meters per second (m/s), although the units are not visible. Assuming standard units, to estimate the distance using four strips, we would divide the 20-second interval into four equal parts, each representing 5 seconds.For each 5-second interval, estimate the average velocity and then multiply by 5 to find the approximate distance traveled in each interval. This is essentially calculating the area of each rectangular strip since Distance = Velocity x Time.To get a better estimate, you might average the velocity at the start and the end of each 5-second interval to find the height of each rectangle (due to the graph's linearity, it will be the velocity at the middle of each interval):Let's denote the velocity at the start of each interval as V1, V2, V3, and V4, and the velocity at the end as V5, V6, V7, and V8 respectively.The approximate distance (D) traveled in 20 seconds using this method would be:D ≈ (V1 + V5) / 2 * 5 + (V2 + V6) / 2 * 5 + (V3 + V7) / 2 * 5 + (V4 + V8) / 2 * 5For part b), since the graph shows a curve that is concave up (the slope is increasing), using four strips will likely result in an underestimate. This is because the rectangles will not cover the entire area under the curve - there will be small triangular areas between the top of each rectangle and the curve that are not being accounted for.
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