Example Question - underestimate and overestimate
Here are examples of questions we've helped users solve.
Estimating Train Distance with Graph
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 * 5 For 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.
Estimation of Train Distance Travelled Over Time
The image displays a graph that shows the velocity of a train over time, and the instructions are to use four strips of equal width to estimate the distance traveled in the first 20 seconds, followed by a question whether this estimation would be an underestimate or overestimate. To solve part (a): 1. The first step is to divide the total time of 20 seconds into four equal time intervals, which means each interval will be 5 seconds long. 2. Determine the approximate velocity of the train at each of the intervals. Since the graph is a curve rather than straight lines, we estimate the average velocity over each interval by taking a value from the graph that seems reasonable, such as the value at the midpoint of each interval. Let's say, for the sake of argument, the velocity values (in meters per second) were approximately determined from the graph as: - At 2.5 seconds: 10 m/s - At 7.5 seconds: 20 m/s - At 12.5 seconds: 30 m/s - At 17.5 seconds: 40 m/s 3. For each interval, calculate the distance traveled by multiplying the average velocity by the interval time (5 seconds): - From 0 to 5 seconds: \( 10 \text{ m/s} \times 5 \text{ s} = 50 \text{ m} \) - From 5 to 10 seconds: \( 20 \text{ m/s} \times 5 \text{ s} = 100 \text{ m} \) - From 10 to 15 seconds: \( 30 \text{ m/s} \times 5 \text{ s} = 150 \text{ m} \) - From 15 to 20 seconds: \( 40 \text{ m/s} \times 5 \text{ s} = 200 \text{ m} \) 4. Add up the distances from each interval to obtain the total estimated distance traveled in the first 20 seconds: - Total distance \( = 50 \text{ m} + 100 \text{ m} + 150 \text{ m} + 200 \text{ m} = 500 \text{ m} \) For part (b): The estimation using this method is an underestimate. This is because the graph shows that the train's velocity is increasing over time, and by taking the average velocity at each midpoint, we under-represent the velocity during the second half of each interval. Therefore, the actual area under the curve (representing the true distance) is larger than the estimated area using the mid-interval velocities for these rectangles. Thus, the distance would be underestimated using this estimation method.
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