What is the y=mx+b equation of a train bowing at 392 km after 2.8 hours, leaving the station at x=0?

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 A train leaves the station at time x=0. Traveling at a constant​ speed, the train travels 392 km in 2.8 hours. Round to the nearest 10 km and the nearest whole hour. Then represent the​ distance, y, the train travels in x hours using an equation. (y=mx+b, preferably)

I just can't figure this out? Im in 8th grade, algebra 1, and just need a simple answer to a simple question.

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Speed of train=392/2.8 km/hr = 140 km/hr

Since the train starts at x=0, the required line passes through the origin. Hence the x intercept, i.e, b is 0. Hence, b=0.

The slope of the line is $\ \frac{rise}{run} = \frac{\Delta y}{\Delta x} = \frac{distance}{time} $ which is the speed.

Note: $\ \Delta y $ means change in y

Hence, m=140

Thus, the line $\ y = 140x $ represents the motion of the train, where the y axis is the distance traveled in km and the x axis is the time in hours.