A train travels along a track and its speed (mph) is given s(t)=29t for the first half hour. Its speed is constant and equal to s(t)=29/2 after the first half hour. t is in hours.

Question

How far (miles) does the train travel in the first hour?

Answer 1

Hey Joe,

The first thing we'll want to do is rewrite the given information in a way that makes sense to us.

1. s(t) = 29*t, 0 ≤ t ≤ .5
2. s(t) = 29/2, .5 < t

The above is just a fancy way of saying:

From t = 0 hrs (the beginning) to t = .5 hrs (the half hour mark), equation "1" holds true.

Then, after t = .5 hrs (the half hour mark) onward, equation "2" holds true.

Now, let's look at the question...

Question: How far (miles) does the train travel in the first hour?

Answer: First, we need to understand the relationship between speed (miles / hour), time (hours), and distance (miles). To do this, let's take a look at their units... this is called dimensional analysis.

If we take units of Speed (which is miles / hour) and multiply it by units of time (which is hours), then we get (miles / hour) * (hours). Since the Speed unit's "hour" denominator cancels out with the time unit's "hour" numerator, the product of the two results in (miles / hour) * (hours) = "miles". Miles is a unit of distance, which is exactly what we're looking for!

Now that we know we need to multiply speed by time to get distance, we can continue onwards. We can think of multiplication as the area under a curve of an xy graph, in this case the graph of speed (y-axis) vs. time (x-axis). Let's see what shaped curve we end up with.

(a) At t = 0 (the beginning), we can use equation "1" to determine the initial speed. The initial speed is s(t=0) = 29*(0) = 0 miles per hour.

(b) At t = .5 (the half hour mark), we can still use equation "1" to determine the speed. The speed is s(t=.5) = 29*(.5) = 14.5 miles per hour.

(c) Right after t = .5 (the half hour mark), we need to switch over to equation "2" to determine the speed. The speed is s(t=.5) = 29/2 = 14.5 miles per hour.

Since the speed values at and right after t = .5 (the half hour mark) are the same, we know that our speed function is continuous. This lets us know that the two curves before and after t=.5 (the half hour mark) are connected.

(d) At t = 1 (the hour mark), we need to continue using equation "2" to determine the speed. The speed is s(t=1) = 29/2 = 14.5 miles per hour. Notice how, with equation "2", t could have been anything and the outcome would still have been the same. This is why they say speed is constant after the first half hour.

Now we know the graph's curve. It is going to:

(a) start at the (x,y) origin (t=0, v=0)

(b) linearly shoot up to (t=.5,v=14.5)

(c-d) remain constant from (t=.5,v=14.5) to (t=1,v=14.5).

From (a) to (b), the shape is a right triangle. This right triangle has a base of (.5-0=.5) and a height of (14.5-0=14.5). The area of a right triangle is .5*base*height. This gives us .5*.5*14.5 = 3.625.

From (c) to (d), the shape is a rectangle. This rectangle has a base of (1-.5=.5) and a height of (14.5-0=14.5). The area of a rectangle is base*height. This gives us .5*14.5 = 7.25.

Add these two areas together (don't forget to include units), and we get a distance of 3.625 miles + 7.25 miles = 10.875 miles as our final answer!

Thanks,

Tristan

Answer 2

Distance will be the area of a triangle with a base of .5 hr and height of 14.5 mph equal to 3.625 miles