What is the total distance traveled?

Question

The instant an army x miles long starts to march, a solder leaves the rear part of the army for the front and reaches it after the army has traveled x miles. The solder then turns back and travels to the rear of the army at the same rate. In terms of x, what is the total distance traveled by the solder?

Answer 1

2.666...x
OR
8x / 3

Assume that the army is four miles long. We can draw a picture to represent this:

(soldier)<________>(front of army)<_______>(turnaround point)
------------- 4 miles-------------------------- 4 miles------------------------

The army must travel 4 miles to reach the turnaround point, and the soldier must travel 8 miles. For the sake of simplicity, lets assume that the first leg of the trip takes one hour. That means that our soldier is traveling at 8mph, and the army is traveling at 4 mph.

We need to figure out how much he must travel to reach the rear of the moving formation, which is 4 miles away when he turns around. Since the soldier and the line are moving towards one another, we can combine their speeds to determine how much time this takes. At 12mph, it takes 1/3 of an hour to close the 4 mile difference. Since our soldier is traveling at 8 mph for 1/3 of an hour, he marches 2 2/3 miles on the second leg of his trip.
Totaling the two legs of the trip, his total distance covered is:
8 + 2 2/3 = 10 2/3 miles.

Since x = 4 in this example, we would divide by four to reduce x to one:
10 2/3 / 4 = 2 2/3x = 8x / 3

Answer 2

If the army of X miles lin length marched X miles, then going from back to front in the same time would require the soldier move at twice the pace of the army, and he would have travelled a distance of 2x.

When the soldier turns back, he's moving at a negative rate compared to the army... he's marching backwards twice as fast as they are moving forward, so his net rate relative to the army is 2X + x = 3X Therefore, in the time he marches backwards to meet the rear, the army would have moved a distance of 1/3X, and he would have move 2/3 X.

So, the answer would be 2x + 2/3 x = 2.66X

Answer 3

In the same duration of time Army travelled X and soldier travelled 2X. So far the soldier has travelled 2X and his speed is twice that of Army.On the return journey the relative speed of soldier is 3X. So he will cover X in 1/3 the time or will travel x/3 to reach the starting point. So total dist.travelled=2X+X/3=5X/3

Answer 4

the guy has travelled 2x miles forward ... since he covered the length of the army (x) plus the distance covered by the army (x)

his speed is therefore = twice the army's

so his return trip will take = x / (x+2x) = 1/3 time taken earlier
in that time he covers 1/3*2x = 0.67x distance

so total distance covered by him = 2x+0.67x = 2.67x

Answer 5

you will desire to have gadgets of m/s, m.s-2 are acceleration gadgets. The gadgets is a mistake. caricature the graph of speed (y-axis) against time (x-axis). Use diagonal or horizontal strains purely whilst becoming to be a member of the factors (no vertical strains). the section under the graph represents the area, so artwork out the 'section' yet splitting the graph into rectangles and triangles. From 0s to 20s section = '0.5 base x top'; represents (a million/2)(20s)(30m/s) = 300m From 20s to 30s section = base x top; represents (30s)(30m/s) = 900m From 50s to 60s section = '0.5 base x top'; represents (a million/2)(10s)(30m/s) = 150m From 70s to 80s section = '0.5 base x top'; represents (a million/2)(10s)(10m/s) = 50m From 80s to 90s section = 'base x top; represents (10s)(10m/s) = 100m From 90s to 100s section = '0.5 base x top'; represents (a million/2)(10s)(10m/s) = 50m entire = 3 hundred + 900 + a hundred and fifty + 50 + one hundred + 50 = 1550m

Answer 6

2x + 2/3x
= 2.67x

Answer 7

3x

Answer 8

my guess is 2.5x