Cars A and B: Rate, Distance and Time Problem

Question

Cars A and B start together on a circular track. Car A travels once around the track at a rate of 30 miles per hour. Car B travels in the opposite direction at 20 miles per hour until it passes car A. From that point, at what speed must car B travel so as to cross the start-finish line at the same time car A does?

Answer 1

1. Distance (D) = Rate (R) * Time (T)
2. For Car  A: 
1. D = 30T    {Eqn 1}
3. For Car B:
1. The Distance D and the time T is the same as for Car A
2. But, unlike Car A, Car B travels at two different rates.  It travels at 20mph for a time T1.  Then it passes Car A and travels at a rate of x for a period of time T2.
3. D = (20T1) + (x * T2)
4. Note that T2 must equal the total time T minus T1, so
5. D = (20T1) + (x * (T - T1))    {Eqn2}
6. Substitute Eqn 1 into Enq 2 :   30T = 20T1 + xT - xT1
7. Rewrite as:  30T = 20T1 + x(T-T1)  {Eqn 3}
4. At what point do the two cars pass one another? 
1. Car A travels 30T1  = D1
2. Car B is traveling in the opposite direction at 20T1 = D2
3. D2 = D - D1 (where D is the total circular length of the track)
4. Substituting, 20T1 = D - 30T1.   Or 50T1 = D   {Eqn 4}
5. Now we can establish a relationship between T and T1 by substituting Eqn1 into Eqn4
1. D = 30T = 50T1
2. 30T = 50T1
3. (30/50)T = T1
4. T1 = (3/5)T   {Eqn 5}
6. Now substitute Eqn 5 into Eqn 3
1. 30T = 20T1 + x (T-T1)
2. 30T = 20(3/5)T+ x (T - 3/5 T)
3. 30T = 12T +  xT(2/5)
4. 18T = (2/5)x
5. (5/2)18 = x
6. 45 mph = x
7. Recall that x is the speed that Car B must travel after it passes Car A. 

Answer 2

give the track a distance such as 60 miles in circumference (you can use 30 or any distance)

car A goes 30 mph and car B goes 20 mph

for every mile car A goes, car B goes 2/3 of a mile

they start together

after car A goes 36 miles car B goes 24 miles in the opposite direction (36+24=60)

car A has to travel 24 more miles to the finish line while car B has to travel 36 more miles to the finish line

distance=rate*time

24=30*t

t=24/30=4/5 of an hour for car A to cross the finish line

car B has 4/5 of an hour to travel the rest of the distance which is 60-24=36 miles

36=r*(4/5)

(5/4)(36)=r*(4/5)(5/4)

45=r

car B must travel at the rate of 45 mph to cross the finish line the same time car A does