Walkway - Fast And Furious II

Bob would be able to reach the end of the walkway at $10$ seconds since $\frac {100 ft}{\frac {(6+4) ft}{s}} = \frac {100 ft}{\frac {10 ft}{s}} = 10 s$ . At that time, Ash would have been on the walkway for $10 + 2 = 12$ seconds and is $12 s \times \frac {6 ft}{s} = 72 ft$ from the start of the walkway, while Chip would have been walking beside the walkway for $10 - 2 = 8$ seconds and is $8 s \times \frac {8 ft}{s} = 64 ft$ from the start.

Now, since Bob is walking $10 \frac {ft}{s}$ , his velocity is at $\frac {(6 - 10) ft}{s} = -4 \frac{ft}{s}$ while Ash and Chip still travel at $6 \frac {ft}{s}$ and $8 \frac {ft}{s}$ , respectively. Thus, from the time Bob decided to turn back, after $t$ seconds, Ash, Bob and Chip would have been $72 + 6t$ , $100 - 4t$ and $64 + 8t$ feet away from the start of the walkway, respectively.

Now, we will determine who was the one at the middle at the time in question. We know that Ash is still in the middle of Bob and Chip at $t = 0$ . If (1) Bob was able to go past Ash before Chip can overtake Ash, then Ash is the one in the middle at the (earliest) instance in question. If (2) Chip was able to overtake Ash before Bob was able to go past Ash, then Chip is the middle person in question. The case that Bob is in the middle of Ash and Chip (note the order) should be overlooked since (1) had to happen first before this does. So should be the case when Bob is in the middle of Chip and Ash because (2) had to happen first before this does. Now, to check this mathematically, we know that Ash and Bob are $28$ feet away from each other at $t = 0$ and in every second, since they are moving in opposite directions, their distance from each other narrows down by $|6 - (-4)| = |6 + 4| = 10$ feet every second and Bob would have gone past Ash sometime before $t = 3$ . Ash and Chip, who are moving in the same direction, are $8$ feet apart and their distance narrows down $|6 - 8| = |-2| = 2$ feet per second which means it would take $4$ seconds before Chip can overtake Ash. Thus, case (1) is our target.

Since Ash is in the middle of Bob and Chip at that time, we will solve in a way that the distance between Bob and Ash is the same as of Ash and Chip, or

$\begin{aligned} (100 - 4t) - (72 + 6t) &= (72 + 6t) - (64 + 8t) \\ 100 - 4t - 72 - 6t &= 72 + 6t - 64 - 8t \\ 28 - 10t &= 8 - 2t \\ 20 &= 8t \\ 2.5 = \frac {20}{8} &= t \end{aligned}$

and at $t = 2.5$ , Ash would have been $72 + 6(2.5) = 72 + 15 = \boxed {87}$ feet from the start of the walkway.

Bob after competing 100 feet will travel back at 4 feet per second (10-6). Ash travels at 6 feet per second and Chip at 8 feet per second.

Ash will again be in the middle because till 16 seconds Chip will not overtake him. Let s be the time at which Ash is in the middle. At that time Chip will be at 8(s-4)= 8s-32

Bob will be at 100- 4(s-12)= 100-4s+48= 148-4s

(8s-32+148-4s)/2= 6s

4s+116= 12s

s=116/8=29/2

Ash will be at 6s= 29/2*6= 87 feet

Because: Time requires for Chip to get the destination is more than Ash, so Ash is between Bob and Chip. Assume X is distance that Ash has gone after returning of Bob. Have equality: $36 - \frac{4 \times X}{3} - \frac{2 \times X}{3} = 2 \times (28 - \frac{5 \times X}{3})$ => X = 87.

Bob covers 100 feet 10 seconds. Therefore:

Ash would have covered 72+6t and Chip would have covered 64+8t in the t seconds Bob walks back at 4 feet per sec which is same as Bob is at a distance of 100-4t from the starting point.

Treating them as ABC (Ash, Bob, Chip), we have 6 sets of (ABC, CBA, BAC, CAB, ACB, BCA). Solving for these equations, we get the time t=2.5 seconds for CAB which gives the only integer answer that is within 100 feet!

The answer is distance travelled by Ash = 72+6x 2.5 = 87

The main thing starts after BOB has reached the last point.....

At that time, ASH reached 72 feet and CHIP has reached 64 feet.

After that BOB moves backward at 10 ft./sec there his speed will be 10-6=4 ft./sec ASH and CHIP are moving in forward direction at 6 and 8 ft per sec. and after that you can see the solution of satyen nabar as i have done the same thing

thanks.