Question

A drunkard is walking along a straight road. He takes 5 steps forward and 3 steps backward and so on. Each step is 1 m long and takes 1 s. There is a pit on the road 11 m away from the starting point. The drunkard will fall into the pit after?

• 1.21s
• 2.29s
• 3.31s
• 4.37s
Option 2:29s
Solution

The drunkard will fall into the pit after 29s. After taking 5 steps forward and 3 steps backward in 8 seconds, the individual is 2 meters distant from his starting position.

It can be written as:

X = 5-3=2 m in 8 sec

He will be 6 m distant from his starting place after 24 seconds.

He advances another 5 m before falling into a pit 11 m away from his starting point.

Total time T = 24 + 5 = 29 sec.

Distance and Displacement

Distance and displacement are two quantities that seem to mean the same thing, but have very different definitions and meanings.

• Distance is a scalar quantity that describes "how much ground an object has covered in its travels."
• Displacement is a vector quantity that expresses "how much an object is displaced". This is the global change in the object's position.

Distance is the total distance traveled by the body over a period of time.

Summary:

A drunkard is walking along a straight road. He takes 5 steps forward and 3 steps backward and so on. Each step is 1 m long and takes 1 s. There is a pit on the road 11 m away from the starting point. The drunkard will fall into the pit after?

A drunkard is walking along a straight road. He takes 5 steps forward and 3 steps backward and so on. Each step is 1 m long and takes 1 s. There is a pit on the road 11 m away from the starting point. The drunkard will fall into the pit after 29s.

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