# Thread: His average speed for the first 4km is what?

1. A man runs a distance of 9km at a constant speed for the first 4km and then 2km/h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4km is what?Total distance = 9kmTotal time = 1 hFor the 1st for 4km, he moved at a constant speed.For the last 5km, his speed was increased by 2km/h.Let the costant speed at 4km be xkm/h.At the last 5km, his speed became (x 2) km/h.The total speed for the whole runs took him[(x 2) x] km/h = (2x 2) km/hspeed s = distance d/time t s = d/tt = 9/(2x 2) 1 = 9/(2x 2)2x 2 = 92x = 7x = 7/2 = 3.5 km/hTherefore the speed for the 4 km = 3.5 km/hSince the speed was increased by 2km/h, that means the speed for the last 5 km/h would become (3.5 2)= 5.5 km/h.But when I checked my answer, my did not seem right. Please I need help. Thank you.

2.

4. Originally Posted by chikis
The total speed for the whole runs took him[(x 2) x] km/h = (2x 2) km/h.
You seem to be missing some plus signs, so it's hard to tell what you are doing, but if you meant the average speed for the whole run is [(x+2)+x], or 2x+2, that's not correct.

5. Originally Posted by chikis
A man runs a distance of 9km at a constant speed for the first 4km and then 2km/h faster for the rest of the distance. The whole run takes him one hour. His average speed for the first 4km is what?Total distance = 9kmTotal time = 1 hFor the 1st for 4km, he moved at a constant speed.For the last 5km, his speed was increased by 2km/h.Let the costant speed at 4km be xkm/h.At the last 5km, his speed became (x 2) km/h.The total speed for the whole runs took him[(x 2) x] km/h = (2x 2) km/hspeed s = distance d/time t s = d/tt = 9/(2x 2) 1 = 9/(2x 2)2x 2 = 92x = 7x = 7/2 = 3.5 km/hTherefore the speed for the 4 km = 3.5 km/hSince the speed was increased by 2km/h, that means the speed for the last 5 km/h would become (3.5 2)= 5.5 km/h.But when I checked my answer, my did not seem right. Please I need help. Thank you.
v=s/t, so t=s/v.

The first part is 4 km at x km/h. So the time that took him, t1 = 4/x hrs.

The second is 5km run at (x+2) km/h. So the time for that part is t2 = 5/(x+2) hrs.

But we also know t1 + t2 = 1 hr.

You should be able to get it from here - and when you do the answer will seem laughably obvious: you might even get it by inspection.

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