A train travelled along a track in 110 minutes , correct to the nearest 5 minutes. Jake finds out that the track is 270 km long. He assumes that the track has been measured I correct to the nearest 10 km. Jake's assumption was wrong. The track was measured correct to the nearest 5 km. Explain how this could affect the average speed of

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A train travelled along a track in 110 minutes , correct to the nearest 5 minutes. Jake finds out that the track is 270 km long. He assumes that the track has been measured I correct to the nearest 10 km. Jake's assumption was wrong. The track was measured correct to the nearest 5 km. Explain how this could affect the average speed of

Answer 1

The change in Track length's accuracy from 10 km to 5 km increases the potential maximum speed of the train. Whereas the minimum potential speed remains the same because the lower bound of length doesn't change. Therefore, this widens the window of error in the calculated average speed of the train. Initially, it varied between 141.78 km/h and 153.49 km/h, whereas now it varies from 141.78 km/h to 156.27 km/h. This gives a more considerable highest possible speed of the train which makes the calculated average speed much more uncertain than before.

Answer 2

## Step1: First, we need to understand how errors in measure can affect calculations. Jake initially thought that the track's length was uncertain to the nearest 10 km and that the time was uncertain by 5 min. According to this, the potential discrepancy in distance is ±5 km, and the time would also have a potential discrepancy of ±2.5 mins. ## Step2: Given his initial assumption, he would perform a worst-case scenario error calculation. Using lower bound and higher bound of distance and time we get two results. For worst situations, low speed corresponds with the highest time and longest distance, and high speed corresponds with the least time and shortest distance: - If the distance is at its lower bound of 265 km and the time is at its upper bound of 112.5 minutes: - The speed would have been \( \frac {265 km}{112.5 min} \times 60 = 141.78 km/h \). - If the track is at its higher bound of 275 km and the time is at its lower value of 107.5 minutes: - The speed would have been \( \frac {275 km}{107.5 min} \times 60 = 153.49 km/h \). Hence, the speed would vary anywhere between 141.78 km/h and 153.49 km/h as per his assumption. ## Step3: However, Jake learned that the track was measured correct to the nearest 5 km. This made a 10 km discrepancy possible. At worst situations: - If the distance is at its lower bound of 265 km and the time is at its upper bound of 112.5 minutes: - The speed would have been \( \frac {265 km}{112.5 min} \times 60 = 141.78 km/h \). - If the track is at its higher bound of 280 km using new information and the time is at its lower value of 107.5 minutes: - The speed would have been \( \frac {280 km}{107.5 min} \times 60 = 156.27 km/h \). This means the train's speed now varies from 141.78 km/h to 156.27 km/h, a potentially faster speed calculated based on the new information about the accuracy of the track measurement. Jake's initially smaller window of variation has expanded based on the new measurement accuracy. #

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