Time and distance word problems
"Time and distance" word problems, often also called "uniform rate" problems, involve something travelling at some fixed steady pace, or else at some average speed.
Whenever you read a problem that involves "how fast", "how far", or "for how long", you should think of the distance equation, d = rt, where d stands for distance, r stands for (constant or average) rate of speed, and t stands for time.
I can ride my bike to work in an hour an a half. If I drive 40mph faster than I bike and it takes me 30 minutes to drive the same distance, how far is it to work?
Using rt_{bike}=rt_{ride}
1.5x=0.5(x+40)
1.5x=0.5x+20
1.5x-0.5x=20
x=20
d=rt, then d=20·1.5=30 miles
Suppose that a car goes from town A to town B at 10km/h and a bus goes from town B to town A at 80km/h. If they both set out at the same time how long will it be before they meet? (Distance from A to B equals to 360 km).
In order to solve this problem we need to find the amount of time the car will take to go from one town to the other at 180km/h (100 + 80).
This can be written as .
Therefore they will meet two hours after they both set out.
At 10 o'clock in the morning a runner begins to run at an average speed of 10km/h. Half an hour later another runner starts out on the same route at an average speed of 12km/h. How long will it be before the second runner catches up with the first one? Over what distance will they have travelled?
The second runner follows the first at an average speed of 12 - 10 = 2km/h. In the first thirty minutes the first runner travels 5km. So we need to calculate how much time is needed to travel 5km at a speed of 2km/h.
That is: .
So the second runner takes two and a half hours, during which time he runs:
d=r·t = 12·2.5 = 30 Km