A drunkard walking in a narrow lane takes 7 steps forward and 4 steps backward followed by same strokes. Each step is 0.5m long and requires 1sec. Find the net displacement of drunkard in three such strokes and average velocity of the drunkard.

Solution

Given, the drunkard walks 7 steps forward and 4 steps backward.

So, the net displacement undertaken by the drunkard = 3 steps in one stroke

Also given, length of 1 step = 0.5 m

Time taken to take 1 step = 1 sec

∴ Displacement in 1 stroke = 3 x 0.5=1.5m

Therefore,

Displacement in 3 strokes = 3 x 1.5 =4.5m

Time taken to walk one step is 1 sec.

Since to complete one stroke, drunkard walks 7 steps forward and 4 steps backward i.e. total of 11 steps, therefore drunkard takes 11s to complete one stroke.

So,

Time taken to complete three strokes, t = 11 x 3 = 33s

Now, average velocity of drunkard is given by,

$v = \frac{4.5}{33} = \frac{15}{110} m / s$

Some More Questions From Motion in Straight Line Chapter

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Motion In Straight Line

A drunkard walking in a narrow lane takes 7 steps forward and 4 steps backward followed by same strokes. Each step is 0.5m long and requires 1sec. Find the net displacement of drunkard in three such strokes and average velocity of the drunkard.

Some More Questions From Motion in Straight Line Chapter

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