Dave drove from town a to town b at an average speed of 70km/h. at the same time, peter drove from town b to town a along the same route. 6 minutes after they had passed each other, peter reached town a while dave was still 2 km away from town b. peter took 24 minutes to travel from town b to town a. a) how far did dave travel before he passed peter? b) find out peter's average speed for the journey

Remark

Every so often, someone comes up with a variation on the most basic physics formula d = r * t

We will need to think up three formulas to solve this. We may not need all three, but we need them to see how the problem works.

Formula one

Where and when do they meet.

The meet after 6 minutes before Peter finished. Peter finished the course after 24 minutes.  Six minutes before that they met. That's 18 after starting they met.

After 18 minutes they completed the course between them

hours to meeting = 18/60 = 0.3 hours

Dave met Peter after a distance of r * t = 70*0.3 = 21 km

Peter had traveled at some speed for 21 minutes = r1 * 0.3 = d1

So the total distance d = is 21 km + r1*0.3  = d

Six minutes later

Peter's Progress

6 minutes later Peter finishes distance D. He does this in 24 minutes.

d = r1 * 24 = r1*24/60 = r1 * 0.4 hours

Dave's Progress

d - 2 = 70 km/hr * 0.4 hr

d - 2 = 28 km

d = 28 + 2 = 30 km

Note this is the total distance that both  Peter and Dave travelled.

Question A

How far had  Dave gone before he met Peter.

Time = 18 minutes = 0.3 of an hour.

r = 70 km / hour

r = 70 km/hour * 0.3 hour

d = 21 km They met after Dave drove 21 km <<< Answer

Question B

Peter's Rate

d = 30 km

t = 24 minutes = 24/60 = 0.4 hours.

r = 30/0.4 = 75 km/hr.  <<<   Peter's Speed Answer

your answer is a

step-by-step explanation:

4*4*4*4=256

maybe showing the graph would

step-by-step explanation:

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