The ‘moving walkway’ is a 300-foot long walkway consisting of a convey

After going through all explanations given in this thread, I was way too confused why the relative speed of Bill is (6-3) 3 feet/sec. Since both Bill and group of people are moving on 3 feet/sec conveyor Belt, 3 feet/sec gets cancelled when Bill moves to catch up the group of people. Below is my explanation of my understanding.

The rate at which Bill moves is (6+3) 9 feet/sec.

The rate at which the group of people moves is 3 feet/sec.

The distance between Bill and group of people is 120 feet.

Since Bill and group of people are moving in same direction, the relative speed will be (9-3) 6 feet/sec
Total Time taken by Bill to cover 120 feet distance is 120/6 = 20 sec.

When Bill makes his move to catch up the group of people, even the group of people moves at 3 feet/sec. Total distance covered by Bill in 20 sec with his rate of 9 feet/sec is 180 feet, that makes remaining distance = 120 feet.

After Bill catches up with group of people he continues at a rate of 3 feet/sec, and time taken to cover that 120 feet at 3 feet/sec will be 40 sec.

Average rate of Bill = Total distance/Total Time = 300/60 = 5 feet/sec.

I did similar to karishma method's

The total distance is 300 foot, it will take 100 seconds if he stands throughout, however he walks for 120 foot.
Hence, he saved time when he walked for 120 foot i.e saved 120/3 = 40 seconds.

Thus, 100 secs - 40 secs = 60 seconds total time taken.

Therefore 300/60 = 5 fps

The concept here tested is "relative speed " and "Average speed"
relative speed of objest and person moving in same direction Sa-Sb
and avg speed= total distance / total time.
1. RS= 6-3=3feet/sec
time take to cover 120 feet to touch crowd= t=d/s
120/3=40 second
but in that 40 seconds crowd would have moved = 40*3 d=s*t= 120feet
so , ben gona catch them at 240 feet coz ben D=speed * time = 40*3=120 feet
now remaing 60 feet he gona travel with crowd @ 3feet/sec. i.e time=d/s
60/3=20 seconds
now , total time for ben = 40+20=60 seconds
and total distance =300 feet avg speed=300/60=5 feet/sec.
hope its clear.

Average Rate of Movement seems like a different way to try to say “average Speed”


Avg Speed = (Total Distance) / (Total Time)


1st) Bill walks with the elevator until he Catches up with the People


Bill’s Effective Speed = (3 + 3) = 6 ft/sec ———as the moving walkways helps push him along and extra 3 ft/sec faster

The Gap Distance that Bill must Catch Up and Close = 120 ft

The group of people are moving with the walkway at 3 ft/sec in the Same Direction

Time = Gap Distance / Relative Speed = 120 / (6 - 3) = 40 seconds

In these 40 seconds, Bill travels: 6 * 40 = 240 feet



The 2nd Part is where Bill moves with the moving walkway and stands still.

60 feet remains

Bill’s Speed at this point = Walkway Speed of 3 ft/sec

Time = 60 / 3 = 20 seconds




Average Speed = (Total Distance of 300 ft) / (Total Time of 40 seconds + 20 seconds)

Avg Speed = 300 / 60 =


5 ft/sec


-E- is the Correct Answer

Posted from my mobile device

I am not sure that i understood the explainations provided. Below is my question:
Bill speed wil be (3ft/sec (walking speed)+ 3ft/sec (conveyor belt speed))= 6ft/sec, now at this rate he will cover 120 ft @ 20sec. Now in 20 sec the group would have traveled additional 60ft i.e 3ft/sec conveyor belt speed* 20sec , so Bill will take 10 more sec, that makes 30sec for Bill to catch with group now the rest 120 (180-120) is what is left will be covered @ 40 sec so the total 300 ft will be covered in 70 sec. I guess my doubt is why is there additional feet which is getting added. Can somebody please explain .

It will take Bill 40 seconds to catch up with the group.

As he’s moving to catch up at 6 ft/sec, the other people are moving also at a rate of 3ft/sec. The entire time that Bill is walking the people are moving with the escalator.

After 40 seconds, eventually Bill’s faster speed will overcome the Distance between him and the group, as well as the fact that the group is constantly moving away from him.


After 40 seconds of walking, Bill meets up and catches the group, he will have traveled 240 feet.

60 feet remains. He “floats” with the escalator for this last 60 feet.


I hope that helps somewhat and didn’t make it anymore confusing?

Posted from my mobile device

Short-cut:

This problem may also be solved with a shortcut. Consider that Bill’s journey will end when the crowd reaches the end of the walkway (as long as he catches up with the crowd before the walkway ends). When he steps on the walkway, the crowd is 180 feet from the end. The walkway travels this distance in (180 feet)/(3 feet per second) = 60 seconds, and Bill’s average rate of movement is (300 feet)/(60 seconds) = 5 feet per

Hi Banuel

For objects on movewalk aways we take relative speed as + for same direction and - for opp?or do we calculate the relative speed from ground(stationary)?

here the relative speed till B reaches group of people can it be determined as b+w(w-walkaway speed)?

Thanks
Nistha

There is a logical way to think about this question

Savg=Dtotal/Ttotal

Since he is going to meet the pple at a certain point, this means that they will reach to the end at the same time! Hence, we can use the time of pple to reach the end of the trip which is t=180f/ 3f/sec = 60sec. Now S=300/60= 5 sec.

Answer is E

I didn't understand why 120/ (6-3)
So this is how I worked towards it:

Imagine a point B
Start of the walkway ------ Point where group is ------ Point B ----- End of the walkway
Distance from where group is to Point B to be X
Therefore: Man's time = group's time
(120+x)/6 = x/3
Which tells us x = 120
Replacing it in the equation
120+120/6 =240/6 =40s

Now we have to look into remaining: 300-240 = 60m
60m/3m/s = 20s

Total : 40+20 = 60s
Therefore 300/60 = 5m/s

brunel: Does this method make sense?