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:
 Samedirection travel
 Roundtrip travel
 Oppositedirection 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 samedirection 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
t1= 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 
t1 
400(t1) 
We equate the distances for the two airplanes and solve for t.
350t = 400(t1)
350t = 400t – 400
350t – 400t = 400
50t=400
t = 8
t1=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 roundtrip 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 oppositedirection 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
 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?
 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?
 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
 This is a samedirection 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(t1/6 ) 
12t = 16(t1/6 )
12t = 16t  2/3
 4t = 2/3
t =2/31/4=2/3
t  1/6 =2/31/6=4/61/6=3/6=1/2
Therefore, Kate catches up with Peter in 1/2 hour or 30 min.
 This is a roundtrip 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 
t1 
250(t1) 
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(t1) to find the distance.
Therefore, Serendra is 1000 miles away.

This is an oppositedirection 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.