- #1
chikis
- 237
- 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 = 9km
Total time = 1 h
For 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/h
speed s = distance d/time t
s = d/t
t = 9/(2x + 2)
1 = 9/(2x + 2)
2x + 2 = 9
2x = 7
x = 7/2 = 3.5 km/h
Therefore the speed for the 4 km = 3.5 km/h
Since 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 did not seem right. Please I need help. Thank you.
Total distance = 9km
Total time = 1 h
For 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/h
speed s = distance d/time t
s = d/t
t = 9/(2x + 2)
1 = 9/(2x + 2)
2x + 2 = 9
2x = 7
x = 7/2 = 3.5 km/h
Therefore the speed for the 4 km = 3.5 km/h
Since 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 did not seem right. Please I need help. Thank you.