# Math: Speed and Distance

Motion Problems:

1. Carlos begins jogging west at 12 km/hr at 10:00 A.M. Denise leaves the same place at 11:00 A.M. and bicycles at 36 km/hr toward Carlos. At what time of day will she overtake him? | Rate| Time| Distance|

Carlos| 12| n| 12n|

Denise| 36| n - 1| 36n - 36|

Note: Denise travels one hour less than Carlos.

When Denise catches up, Carlos and Denise will have travelled equal distances.

12n = 36n - 36

-24n = -36

n = 1.5

Answer: Denise overtakes Carlos at 11:30 A.M.

2. It takes a plane 40 min. longer to fly from Boston to Los Angeles at 525 mph than it does to return at 600 mph. How far apart are the cities?

| Rate| Time| Distance|

Boston to LA| 525| n + 2/3| 525n + 350|

Return trip| 600| n| 600n|

Assume the flight paths are of equal length:

Distance from Boston to LA) = (Return trip distance)

525n + 350 = 600n

350 = 75n

14/3 = n

Using this value of n, we can compute the length of either trip:

Return trip = 600(14/3) = 2800 miles.

ANSWER: The cities are 2800 miles apart.

3. One hour after Yolanda started walking from X to Y, a distance of 45 miles; Bob started walking along the same road from Y to X. If Yolanda’s walking rate was 3 miles per hour and Bob’s was 4 miles per hour, how many miles has Bob walked when they met?

We can make a quick diagram of the problem.

One way to solve this problem is with a table. We want to know the distance Bob walked. The distance Bob walked is our variable. Let x = the distance Bob walked.

Together Yolanda and Bob walked 45 miles, so Yolanda walked 45 -x miles when they met. We are given the rate at which both of them walk. We are told Yolanda walked 1 more hour than Bob. If Bob walked t, Yolanda walked t + 1. The results are summarized in the table below.

We get two equations:

i) x = 4tii) 45 - x = 3t + 3

We can substitute the...

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