Question

Two cars leave towns 800 kilometers apart at the same time and travel toward each other. One car's rate is 10 kilometers per hour less than the other's. If theymeet in 5 hours, what is the rate of the slower car?Do not do any rounding

Answer 1

Formulating a system of equation for solving this exercise, we have:

Let A represent the distance traveled by the first car.

Let B represent the distance traveled by the second car.

800 will be the initial distance of car B.

0 will be the initial distance of car A.

Let V represent the velocity of car A.

Let Z represent the velocity of car B.

Let t represent time in hours. It is equal to 5.

A= 0 + V(t) Equation 1

B= 800 - Z(t) Equation 2

Solving the system of equations we have:

`\begin{gathered} (A)/(t)=V\text{ (Dividing by t on both sides of the equation 1)} \\ B-800=-Z(t)\text{ (Subtracting 800 from both sides of the equation 2)} \\ (B-800)/(-t)=Z\text{ (Dividing the equation 2 by }-t) \end{gathered}`
`\begin{gathered} (A)/(5)=V\text{ (Replacing t=5 in the equation 1)} \\ (B-800)/(-5)=Z(\text{ (Replacing t=5 in the equation 1)} \\ (A)/(5)-(B-800)/(-5)=V-Z\text{ (Subtracting the equations)} \\ (A)/(5)-(B-800)/(-5)=10\text{ (Let us say that V is greater than Z and the difference between them is 10)} \\ (A)/(5)-(A-800)/(-5)=10\text{ (Given that they met at the same distance, then A=B)} \\ -A-(A-800)=-50\text{ (Multiplying on both sides of the equation by }-5) \\ -A-A+800=-50\text{ (Distributing)} \\ -2A+800=-50\text{ (Subtracting like terms)} \\ -2A=-850\text{ (Subtracting 800 from both sides of the equation)} \\ A=425\text{ (Dividing by 2 on both sides of the equation}) \\ \text{They met at 425 m.} \end{gathered}`
`\begin{gathered} \text{ Replacing A=425 in the equation 1} \\ (425)/(5)=V=85 \\ \text{ The difference between V and Z is 10,so Z is equal to 75.} \\ \text{The answer is 75 km/h} \end{gathered}`

The answer is 75 km/h