In the formula d = rt, the time t varies inversely with the rate r. A student running at 5 miles per hour runs one lap around the school campus in 8 minutes. If a second student takes 10 minutes to run one lap around the school, how fast is she running?
Plug the numbers into d = rt and solve for distance, then redo and calculate r for the second student. Post your work if you get stuck.
To find the rate at which the second student is running, we can use the formula d = rt. We know that the distance around the school campus is the same for both students since they are running one lap. Let's denote the unknown rate of the second student as r2.
For the first student, we are given that their rate (r1) is 5 miles per hour and the time (t1) is 8 minutes. However, we need to convert the time to hours in order to have consistent units. Since there are 60 minutes in an hour, t1 = 8 minutes / 60 minutes/hour = 0.1333 hours.
For the second student, we are given the time (t2) as 10 minutes. Similarly, we need to convert this to hours. Therefore, t2 = 10 minutes / 60 minutes/hour = 0.1667 hours.
Since the distance (d) is the same for both students, we can set up the equation:
d = r1 * t1 = r2 * t2
Substituting the values we know, we get:
5 miles/hour * 0.1333 hours = r2 * 0.1667 hours
Simplifying the equation:
0.6665 miles = 0.1667 r2
Now, to solve for r2, we divide both sides of the equation by 0.1667:
r2 = 0.6665 miles / 0.1667 hours
Calculating this, we find:
r2 ≈ 4 miles per hour
Therefore, the second student is running at a rate of approximately 4 miles per hour.