Question

143.(Continuation) Using rate, time, and distance as the context, write a problem that could be solved using the equation in part (b) of the previous problem

Answer 1

We have to write a problem using rate, time and distance as the context that can be solved with the equation given.

We can start by relating rate (speed), time and distance:

`v=(d)/(t)`

this equation represents the average speed, that can be calculated as the quotient between the distance and time.

We can also write, derived from the previous equation:

`\begin{gathered} t=(d)/(v) \\ d=v\cdot t \end{gathered}`

The equation t = d/v can be useful in this case, as we can let x represent distance, 1/2 represent total time and 6 and 4 represent speed in different parts of a path.

We then can write something like:

`t=t_1+t_2=(d_1)/(v_1)+(d_2)/(v_2)`

For example a trip between places A and C. There is a place between A and C, called B.

We walk from A to B at 6 km/h and then from B to C at 4 km/h. It takes half an hour (1/2 hour) to get from A to C.

We also know that the distance from B to C is one kilometer less than the distance from

A to B.

The question is: what is the distance from A to B?

Let x be the distance from A to B.

We can write this problem as:

`\begin{gathered} t=t_(AB)+t_(BC)=(d_(AB))/(v_(AB))+(d_(BC))/(v_(BC))_{} \\ (1)/(2)=(d_(AB))/(6)+(d_(BC))/(4) \end{gathered}`

As the distance from B to C is one km less than from A to B we can write:

`\begin{gathered} d_(AB)=x \\ d_(BC)=d_(AB)-1=x-1 \end{gathered}`

replacing in the equation, we get:

`(1)/(2)=(x)/(6)+(x-1)/(4)`

that is equivalent to the equation given.

Answer:

The problem can be stated as:

"We have to go from A to C. There is a place between A and C, called B.

We walk from A to B at 6 km/h and then from B to C at 4 km/h. It takes half an hour (1/2 hour) to get from A to C.

We also know that the distance from B to C is one kilometer less than the distance from A to B.

The question is: what is the distance from A to B?"