Question

Please see attached picture to understand where I need help

Answer 1

We are given that two trains start a distance of 992 km apart. If they meet after 4 hours, this means that the sum of the distances of both trains after 4 hours traveling must be 992 km. This is written mathematically as:

`d_a+d_b=992`

Now, the distance traveled by an object is given by the following relationship:

`\begin{gathered} d_a=r_at \\ d_b=r_bt \end{gathered}`

Where:

`\begin{gathered} r_a,r_b=\text{ rates of the trains} \\ t=\text{ time} \end{gathered}`

Now, we substitute the values in the equation for the total distance:

`r_at+r_bt=992`

Now, the time is 4 hours, therefore, we have:

`r_a(4)+r_b(4)=992`

Now, we are also given that one train travels 20 km/h slower than the other. If a is the slower train then we have:

`r_a=r_b-20`

Substituting we get:

`4(r_b-20)+4r_b=992`

Now, we divide both sides by 4:

`r_b-20+r_b=(992)/(4)`

Solving the operations:

`r_b-20+r_b=248`

Now, we add like terms:

`2r_b-20=248`

Now, we add 20 to both sides:

`\begin{gathered} 2r_b=248+20 \\ 2r_b=268 \end{gathered}`

Now, we divide both sides by 2:

`r_b=(268)/(2)=134`

Therefore the fastest train has a rate of 134 km/h. Now, we determine the rate of the slower train:

`\begin{gathered} r_a=134-20 \\ r_a=114 \end{gathered}`

Therefore, the slower train travels ar 114 km/h