Question

6. A is 60 miles from B. An automobile at A starts for B at the rate of 20 miles an hour atthe same time that an automobile at B starts for A at the rate of 25 miles an hour. Howlong will it be before the automobiles meet?RateTimeDistanceRyanCastel

Answer 1

6. A is 60 miles from B. An automobile at A starts for B at the rate of 20 miles an hour at

the same time that an automobile at B starts for A at the rate of 25 miles an hour. How

long will it be before the automobiles meet?

Rate

Time

Distance

Ryan

Castel

Step 1

Assume they are were they meet

Let

Ryan goes from A to B

rate = 20 miles per hour

time = unknow (t), this time is the same for the two automobile

distance1=unknown( x)

Castel goes from B to A

rate2= 25 miles per hour

time= unknown(t)

distance2 =unknown(y)

we also know that distance from A to b is 60 miles, so

`\begin{gathered} \text{distance}1+\text{distance}2=60\text{ miles} \\ x+y=600 \\ y=600-x \end{gathered}`

Step 2

make the equations

for Ryan

`\begin{gathered} \text{rate1}=\frac{dis\tan ce}{\text{time}} \\ 20=(x)/(t) \\ t=(x)/(20) \end{gathered}`

For Castel

`\begin{gathered} \text{rate}=\text{ }\frac{dis\tan ce}{\text{time}} \\ \text{25}=(y)/(t)=(600-x)/(t) \\ 25\cdot t=600-x \\ t=(600-x)/(25) \end{gathered}`

Now, the time is the same

`\begin{gathered} t=t \\ (x)/(20)=(600-x)/(25) \end{gathered}`

solve for x

`\begin{gathered} 25x=20(600-x) \\ 25x=12000-20x \\ 25x+20x=12000 \\ 45x=12000 \\ x=(12000)/(45) \\ x=266.66\text{ miles} \end{gathered}`

now, with the value of x, replace it to find t

`\begin{gathered} t=(x)/(20) \\ t=(266.66)/(20) \\ t=13.33\text{ hours} \end{gathered}`

finally, replace the value of x to find y