Question

Elena bikes 20 minutes each day for exercise. Write an equation to describe the relationship between her distance in miles, D, and her biking speed, in miles per hour, when she bikes: a. At a constant speed of 13 miles per hour for the entire 20 minutes b. At a constant speed of 15 minutes per hour for the first 5 minutes, then at 12 miles per hour for the last 15 minutes c. At a constant speed of M miles per hour for the first 5 minutes, then at N miles per hour for the last 15 minutes

Answer 1

Answer:

The equation to describe the relationship between Elena distance:

a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

Solution:

Given, Elena bikes 20 minutes each day for exercise.  

We have to write an equation to describe the relationship between her distance in miles, D, and her biking speed, in miles per hour,  

We know that, distance travelled = speed $\times$ time

a. At a constant speed of 13 miles per hour for the entire 20 minutes  

Her speed is 13 miles per hour.  

Then, distance D miles = 13 miles per hour $\times$ 20 minutes

$\mathrm{D}=13 \text { miles per hour } \times \frac{20}{60} \text { hours } \rightarrow \mathrm{d}=13 \times \frac{1}{3} \rightarrow \mathrm{d}=4.34 \text { miles approximately. }$

b. At a constant speed of 15 minutes per hour for the first 5 minutes, then at 12 miles per hour for the last 15 minutes  

Now, total distance travelled = distance travelled with 15 mph + distance travelled with 12 mph

$\begin{array}{l}{\mathrm{D}=15 \mathrm{mph} \times 5 \text { minutes }+12 \mathrm{mph} \times 15 \text { minutes }} \\\\ {\mathrm{D}=15 \mathrm{mph} \times \frac{5}{60} \text { hours }+12 \mathrm{mph} \times \frac{15}{60} \text { minutes }} \\\\ {\mathrm{D}=15 \times \frac{1}{12}+12 \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{5}{4}+3 \rightarrow \mathrm{D}=3+1.25 \rightarrow \mathrm{D}=4.25 \mathrm{miles}}\end{array}$

c. At a constant speed of M miles per hour for the first 5 minutes, then at N miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with M mph + distance travelled with N mph

$D = M mph \times 5 minutes + N mph \times 15 minutes$

$\mathrm{D}=\mathrm{M} \mathrm{mph} \times \frac{5}{60} \text { hours }+\mathrm{N} \mathrm{mph} \times \frac{15}{60} \text { minutes }$

$\mathrm{D}=\mathrm{M} \times \frac{1}{12}+\mathrm{N} \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{N}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{3 N}{12} \rightarrow 12 \mathrm{D}=\mathrm{M}+3 \mathrm{N} \text { miles }$

Hence, a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

Answer 2

Speed is the rate of change of distance over time.

• The distance is 4.33 miles
• The distance is 4.25 miles
• The expression for distance is $\mathbf{D = \frac{M}{12} + \frac{N}{4}}$

The given parameter is:

$\mathbf{t = 20\ mins}$

(a) The distance, when her speed is 13mph

Distance (D) is calculated using:

$\mathbf{D = Speed \times Time}$

So, we have:

$\mathbf{D = 13mph \times 20\ mins}$

Convert time to hours

$\mathbf{D = 13mph \times \frac{1}{3}\ h}$

$\mathbf{D = 4.33\ miles}$

The distance is 4.33 miles

(b) The distance, when her speed is 15mph for the first 5 minutes and 12mph, for the last 15 minutes

Distance (D) is calculated using:

$\mathbf{D = \sum Speed \times Time}$

So, we have:

$\mathbf{D = 15mph \times 5\ mins + 12mph \times 15\ mins}$

Convert time to hours

$\mathbf{D = 15mph \times \frac{1}{12}\ h + 12mph \times \frac{1}{4}\ h}$

$\mathbf{D = 1.25\ miles + 3\ miles}$

$\mathbf{D = 4.25\ miles}$

The distance is 4.25 miles

(c) The distance, when her speed is M for the first 5 minutes and N, for the last 15 minutes

Distance (D) is calculated using:

$\mathbf{D = \sum Speed \times Time}$

So, we have:

$\mathbf{D =M \times 5\ mins + N \times 15\ mins}$

Convert time to hours

$\mathbf{D = M\times \frac{1}{12}\ h + N\times \frac{1}{4}\ h}$

$\mathbf{D = \frac{M}{12} + \frac{N}{4}}$

The expression for distance is $\mathbf{D = \frac{M}{12} + \frac{N}{4}}$

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