Question

mary can jog twice as fast as she can walk. she was able to jog the first 7 miles to her grandmothers house, tired out and walked remaining 1 mile. total trip took 1.5 hours, what was her average jogging speed

Answer 1

6 miles per hour

Step-by-step explanation

Step 1

Let

x,the speed when she walks

then

2x is the speed when she jogs

total distance= 7+1= 8 miles

time joggin= time1

time walking=time2

total time = 1.5 hours

time1+time2 = 1.5

for the first 7 miles

`\begin{gathered} \text{speed(x)}=\frac{dis\tan ce\text{ }}{\text{time}} \\ \text{for the first 7 miles}(\text{joggin)} \\ \text{speed}=\frac{7\text{ miles}}{\text{time}1} \\ 2x=\frac{7}{\text{time 1}}\text{ equation(1)} \\ \\ \end{gathered}`

for the last mile

`\begin{gathered} x=\frac{1\text{ miles}}{\text{time2}} \\ x=\frac{1}{\text{time}2}\text{ Equation (2)} \end{gathered}`

Step 2

the total time is 1.5

`\text{time 1+time2=1.5 equation (3)}`

using equation (1) ,(2) and (3) find time 1 and time 2

a) replace the value for x from equation (2) in equation(1)

`\begin{gathered} 2(\frac{1}{\text{time}2})=\frac{7}{\text{time 1}} \\ \\ \frac{2}{\text{time}2}=\frac{7}{\text{time 1}} \\ \text{time}1=(7\cdot time2)/(2)\text{ Equation(4)} \end{gathered}`

replace this value inequation (3)

`\begin{gathered} \text{time 1+time2=1.5 equation (3)} \\ (7\cdot time2)/(2)\text{+time2=1.5 } \\ \text{time}2((7)/(2)+1)=1.5 \\ \text{time}2(4.5)=1.5 \\ \text{time}2\text{ = }(1.5)/(4.5) \\ \text{time}2=0.33\text{ Hous} \end{gathered}`

replace this value in equation (3) to find time 1

`\begin{gathered} \text{time 1+time2=1.5} \\ \text{time 1= 1.5-time2} \\ \text{time}1\text{ =1.5-0.33} \\ \text{time}1=1.16666\text{ hours} \end{gathered}`

Step 3

finally the average joggin speed is

`\text{average speed= }\frac{dis\tan ce}{\text{time}}=\frac{7\text{ miles}}{1.66\text{ hours}}=6\text{ miles per hour}`

I hope this helps you