The distance between A and C need not be given, it is redundant.PollutedS wrote:Hi! please help

In a time trial a cyclist wanted to reach an average of 25km per hour between two towns A and C which are 10km apart. A village, B, is sited exactly halfway between A and C and is reached after a long climb up from A. When the cyclist had climbed up to B she calculated that her average speed so far was only 20km per hour. How fast must she ride on the descent from B to C if she is to attain the overall average speed of 25km per hour?

Since the distance is given it is easier to work....

Average speed = total distance/total time.

so, total time taken for the trial = 10/25 h = 0.4h

Time from A to B

= distance AB/average speed from A to B

= 5/20 h

= 0.25h

Therefore, the time remaining to complete B to C

= 0.4h - 0.25h

= 0.15h

Average speed for B to C

= distance BC /time taken for BC

= 5/0.15 km/h

= 33 1/3 km/h

**suppose the distance A to C is not given, then we can make use of speed ratio/ time ratio to work ie ratio of the speeds = inverse ratio of the time taken ...**

for the SAME distance D,

S1 : S2 = T2 : T1

where T1 is the time @speed S1 and S2 is the time taken @speed S2 to cover the SAME distance D.

for distance AB,

ratio of target average speed : achieved speed

= 25 : 20

= 5 : 4

this means the corresponding time taken to cover AB (= inverse of the speed ratio)

@25km/h : @20km/h = 4u : 5u

since distance BC = distance AB, then

time taken to cover BC @25km/h = 4u

--> total time for distance AC = 8u @25km/h

so, to achieve target average speed of 25km/h from A to C, then remaining time for distance BC = 8u - 5u = 3u

for distance BC,

time taken @required speed : time taken @target speed of 25 km/h

= 3u : 4u

so,

required speed : target speed @25km/h

= 4 : 3 (inverse of the time ratio)

or required speed = target speed x 4/3 = 25 x 4/3 km/h = 33 1/3 km/h

cheers.