Distance Rate Time

An object that moves at a constant rate is said to be in uniform motion. The relationship among distance d, rate r, and time t is d=rt. The uniform motion problems to be discussed will be of three types:

  1. Same-direction travel
  2. Round-trip travel
  3. Opposite-direction travel

Example 1

Airplane A takes off from the Changi Airport at 10:00 AM and travels at the rate of 350 mi/h. One hour later, Airplane B takes off from the same airport following the same flight path at 400 mi/h. In how many hours will the Airplane B catch up with the Airplane A?

Solution:

This is a same-direction travel (type 1). The rates and time for the two airplanes are different, however, the distances they travel (from Changi Airport to the point where Airport B catches up with Airplane A) are the same.

Let t=time the first airplane travels

t-1= time the second airplane travels

We make a table to summarize all the information.

 

Rate

Time

Distance=rate x time

Airplane A

350 mi/h

t

350t

Airplane B

400 mi/h

t-1

400(t-1)

We equate the distances for the two airplanes and solve for t.

350t = 400(t-1)

350t = 400t – 400

350t – 400t = -400

-50t=-400

t = 8

t-1=7

Therefore, Airplane B will catch up with Airplane A in 7 hours.

Example 2

William drives to Ayala Mall to pick up his mother at an average of 15 mi/h. On their way home, because of better traffic condition, William averages 30 mi/h. If the total travel time is 2 hours, how long does it take him to drive to Ayala Mall?

Solution:

This is a round-trip travel (type 2). The distances to and from Ayala Mall are equal.

Let t = time of William’s drive to Ayala Mall

2 – t = time of William’s drive back home

We make a table to summarize all the information.

 

Rate

Time

Distance=rate x time

Travel to Ayala Mall

15 mi/h

t

15t

Travel home

30 mi/h

2 - t

30(2 – t)

We equate the distances for the two trips and solve for t.

15t = 30(2 – t)

15t = 60–30t

45t = 60

t = 60/45 or 1 1/3

Therefore, it took William 1 1/3 hours or 1 h 20 min to drive to Ayala Mall.

Example 3

Johanna and Luke leave their office traveling in opposite directions on a straight road. Luke drives 10 mi/h faster than Johanna. After 1.5 hours, they are 120 miles apart. Find the rate of each one.

Solution:

This is an opposite-direction travel (type 3). The time of travel (1.5 hours) is the same for the two persons. The sum of their distances traveled is 120 miles.

Let r = rate of Johanna

r+10 = rate of Luke

We make a table to summarize all the information.

 

Rate

Time

Distance=rate x time

Johanna

r

1.5 h

1.5r

Luke

r+10

1.5 h

1.5(r+10)

Set the sum of the distances equal to 120 miles.

1.5r + 1.5(r+10) = 120

1.5r + 1.5r + 15 = 120

3r + 15 = 120

3r = 105

r = 35

Johanna was driving at 35 mi/h while Luke was driving at 45 mi/h.

Try these problems

QUESTIONS

  1. Peter leaves school on his bike at 4:00 PM, going at 12 mi/h. Kate leaves the same school 10 min later, going at 16 mi/h in the same direction. How long will it take Kate to catch up with Peter?

  2. It takes 1 hour longer to fly to Serendra at 200 mi/h than it does to return at 250 mi/h. How far away is Serendra?

  3. At 8:00 AM, a car leaves Wilcon Garage at the rate of 60 mi/h. At the same time, another car leaves the same garage at the rate of 50 mi/h in the opposite direction. At what time will the cars be 165 miles apart?

ANSWERS

  1. This is a same-direction travel (type 1). The rates and time for Peter and Kate are different but the distances (up to the point that Kate catches up with Peter) are the same.
    Let t=time of Peter’s travel
    t – 1/6 = time of Kate’s travel (10min is 1/6 hour)
    We make a table to summarize all the information.

     

    Rate

    Time

    Distance=rate x time

    Peter

    12 mi/h

    t

    12t

    Kate

    16 mi/h

    t - 1/6

    16(t-1/6 )

    12t = 16(t-1/6 )
    12t = 16t - 2/3
    - 4t = -2/3
    t =2/3-1/4=2/3
    t - 1/6 =2/3-1/6=4/6-1/6=3/6=1/2
    Therefore, Kate catches up with Peter in 1/2 hour or 30 min.
  2. This is a round-trip travel (type 2). The distances to and from Serendra are equal.
    Let t = time to fly to Serendra
    t – 1 = time for return flight
    We make a table to summarize all the information.

     

    Rate

    Time

    Distance=rate x time

    Flight to Serendra

    200 mi/h

    t

    200t

    Return flight

    250 mi/h

    t-1

    250(t-1)

    We equate the distances for the two trips and solve for t.
    200t = 250(t– 1)
    200t = 250t - 250
    -50t = -250
    t = 5
    Substitute this in 200t or 250(t-1) to find the distance.
    Therefore, Serendra is 1000 miles away.
  3. This is an opposite-direction travel (type 3). The time of travel (unknown) is the same for the two cars. The sum of the distances traveled is 165 miles.
    Let t = time of travel for each of the two cars
    We make a table to summarize all the information.

     

    Rate

    Time

    Distance=rate x time

    First car

    60 mi/h

    t

    60t

    Second car

    50 mi/h

    t

    50t

    Set the sum of the distances equal to 165 miles.
    60t + 50t = 165
    110t = 165
    t = 1.5
    The two cars will be 165 mi apart after 1.5 hours.