Question

4.8 rate - time - distance problems
objective: to solve some word problems involving uniform motion.
warm - up: motion in the opposite direction. mary beth and michael leave school traveling in opposite directions. michael is walking and mary beth is biking, averaging 6 km/h more than michael. if they are 18 km apart after 1.5 h, what is the rate of each?

	rate	time	distance
michael

Explanation:

Step1: Let Michael's rate be $x$ km/h.

Then Mary Beth's rate is $(x + 6)$ km/h.

Step2: Use the formula $d=rt$ (distance = rate×time).

The distance Michael travels in 1.5 h is $1.5x$ km, and the distance Mary Beth travels in 1.5 h is $1.5(x + 6)$ km.

Step3: Since they are moving in opposite - directions, the sum of their distances is 18 km.

Set up the equation $1.5x+1.5(x + 6)=18$.

Step4: Expand the equation.

$1.5x+1.5x+9 = 18$.

Step5: Combine like - terms.

$3x+9 = 18$.

Step6: Subtract 9 from both sides.

$3x=18 - 9$, so $3x = 9$.

Step7: Solve for $x$.

Divide both sides by 3: $x=\frac{9}{3}=3$.

Answer:

Michael's rate: 3 km/h
Mary Beth's rate: $3 + 6=9$ km/h