solve using rate, time, and distance using linear equations

Question

Two cyclists leave towns
128km
apart at the same time and travel toward each other. One cyclist travels
4

/kmh
slower than the other. If they meet in
2
hours, what is the rate of each cyclist?

Answer 1

Any ideas, Ananinah?

 

Suppose the rate of the slower cyclist is X km/h, so the rate of the faster cyclist is (X +4) km/h.

And both of them traveled 2 hours, So the distance of the slower cyclist is 2 X km, the distance of the faster cyclist is 2(X + 4) km. Remember the distance formula D = Rate X Time.

 

Total distance both of them traveled is 128 km/h.

 

So, we can set up the equation:

 

2 X + 2 (X + 4) = 128

2 X + 2 X + 8 = 128

4 X = 120

X = 30 (km/h)--- The rate of the slower cyclist

X + 4 = 34 (km/h)--- The rate of the faster cyclist

 

 

Answer 2

Hi Aaminah,

 

Let one of the cyclists' rate be x km/h. Then, the other one will have a rate of (x-4) km/h.

 

They travel for two hours, so the distances they cover are (using distance = times*rate)

2x and 2(x - 4) km. The sum of these two distances must equal 128 km. Then, we can write

2x + 2(x - 4) = 128

2x + 2x - 8 = 128

4x - 8 = 128

4x = 136

x = 34 km/h (rate of the faster cyclist)

34 - 4 = 30 km/h (rate of  the other cyclist)

Answer 3

We can make use of the formula Distance = Rate x Time

 

For cyclist 1, let his distance traveled = x

For cyclist 2, his distance traveled = 128 - x

 

Now, let cyclist 1 rate = r

Then cyclist 2 rate = r - 4

 

The time traveled by both cyclists is the same: 2 hours. We now have a system of two equations

 

x = 2r

128 - x = (r - 4)(2)

 

Let's use the Substitution Method to solve this system.

 

128 - 2r = 2(r - 4)

128 - 2r = 2r - 8

128 = 4r - 8

136 = 4r

r = 34

 

Cyclist 1 travels at a rate of 34 km/h, and cyclist 2 travels at a rate of 34 - 4 = 30 km/h