Step-by-step explanation:
In this problem, the student is running at a constant speed, which is the center of the problem, because that allow us to use the mathematical relation between speed, distance and time, in a constant movement. It's important to say that constant movement refers to the absence of acceleration, that it, the speed doesn't change. So, the general relation is set by the equation:
<h3>(a) at a constant speed of 4 miles per hour for the entire 30 minutes.</h3>
First, all magnitudes must have same units, there cannot be hours and minutes at the same time, so we'll transform 30 minutes in to hours. For that purpose, we know that 1 hour equals 60 minutes. Therefore, 30 minutes is half an hour, or 0.5 hours, then:
So, the student runs 2 miles.
<h3>(b) at a constant speed of 5 miles per hour the first 20 minutes, and then at 4 miles per hour the last 10 minutes.</h3>
In this case, we have two movements.
- First movement: 5 miles/hour in 20 minutes.
- Second movement: 4 miles/hour in 10 minutes.
We need to transform minutes into hours:
Now, we can the expressions and calculate the distances:
First movement:
Second movement:
So, the student ran 2.33 miles in total.
<h3>(c) at a constant speed of 6 miles per hour the first 15 minutes, and then at 5.5 miles per hour for the remaining 15 minutes.</h3>
Same processes.
First movement:
Second movement:
Therefore, the student ran 2.88 miles in total.
<h3>(d) at a constant speed of <em>a</em> miles per hour the first 6 minutes, and then at 6.5 miles per hour for the remaining 24 minutes.</h3>
First movement:
Second movement:
The student ran (0.10a + 2.6) miles.
<h3>(e) at a constant speed of 5.4 miles per hour for <em>m</em> minutes, and then at b miles per hour for <em>n </em>minutes.</h3>
First movement:
Second movement:
Therefore, the student ran (0.09m + (bn/60)) miles.