Question

A student on the cross-country team runs 30 minutes a day as a part of her training,

Answer 1

Answer: D = time *speed

Step-by-step explanation:

A. 1/2 *4

B. 1/3*5 + 1/6*4

C. 1/4*6 + 1/4*5.5

D. 1/10*1 + 9/10* 6.5

Answer 2

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:

$d = vt$

(a) at a constant speed of 4 miles per hour for the entire 30 minutes.

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:

$d = (4miles/hour)(0.5hours)=2miles$

So, the student runs 2 miles.

(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.

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:

$20min.\frac{1hr}{60min} =0.33hr.\\10min\frac{1hr}{60min}=0.17hr$

Now, we can the expressions and calculate the distances:

First movement:

$d = (5miles/hr)(0.33hr)=1.65miles.$

Second movement:

$d = (4miles/hr)(0.17hr)=0.68miles.$

So, the student ran 2.33 miles in total.

(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.

Same processes.

$15min.\frac{1hr}{60min} =0.25hr$

First movement: $d = (6miles/hr)(0.25hr)=1.5miles.$

Second movement: $d = (5.5miles/hr)(0.25hr)=1.38miles.$

Therefore,  the student ran 2.88 miles in total.

(d) at a constant speed of a miles per hour the first 6 minutes, and then at 6.5 miles per hour for the remaining 24 minutes.

$6min.\frac{1hr}{60min} =0.10hr\\24min.\frac{1hr}{60min}=0.40hr$

First movement: $d = (a miles/hr)(0.10hr)=0.10a (miles).$

Second movement: $d = (6.5miles/hr)(0.40hr)=2.6 miles.$

The student ran (0.10a + 2.6) miles.

(e) at a constant speed of 5.4 miles per hour for m minutes, and then at b miles per hour for n minutes​.

$m (minutes).\frac{1hr}{60min}= \frac{m}{60} hr\\n (minutes).\frac{1hr}{60min}= \frac{n}{60} hr$

First movement: $d = (5.4 miles/hr)(\frac{m}{60} hr)=5.4\frac{m}{60} (miles).$

Second movement:  $d = (b miles/hr)(\frac{n}{60} hr)=b \frac{n}{60} (miles).$

Therefore, the student ran (0.09m + (bn/60)) miles.