Question

A man first walks 2s at 2m/s towards north then 3s at 1m/s towards west .Find average speed and average velocity?​

Answer 1

GIVEN

A man first walks 2 s at 2 m/s towards north then 3 s at 1 m/s towards west.

                   

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TO FIND

The average speed and average velocity.

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WE MUST KNOW

◙ Distance = Speed × Time

◙ Displacement = Velocity × Time

◙ Average speed = Total distance covered/Total time taken

◙ Average velocity = Overall displacement/Total time taken.

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SOLUTION

Case - 1 (When the man walks towards North)

Time, t = 2 s

Speed, v = 2 m/s

Distance = t × v

⇒Distance = 2 × 2 = 4 m

⇒Distance = 4 m

Case - 2 (When the man walks towards West)

Time, t = 3 s

Speed, v = 1 m/s

Distance = t × v

⇒Distance = 3 × 1

⇒Distance = 3 m

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For the overall displacement :-

AB = 4 m

BC = 3 m

AC = ?

Using Pythagoras Theorem,

AC² = AB² + BC²

⇒AC² = (4)² + (3)²

⇒AC = √(16 + 9)

⇒AC = √25

⇒AC = 5 m

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◙ Total distance covered = 4 m + 3 m = 7 m

◙ Total time taken = 2 s + 3 s = 5 s

Average speed = 7/5

⇒Average speed = 1.4 m/s

◙ Overall displacement = 5 m

◙ Total time taken = 5 s

Average velocity = 5/5

⇒Average velocity = 1 m/s

Answer 2

Answer:

Average speed = 1.4 m/s

Average velocity = 1 m/s

Explanation:

Given:

• A man walks with 2 m/s for 2 seconds towards north
• He moves with 1 m/s for 3 seconds towards west

To find:

• Average speed
• Average velocity

Average speed = $\frac{Total \ distance}{Total \ time}$

total distance = 2×2 + 1×3

Total distance =4+3

Total distance = 7 metres

Total time = 2+3 = 5 seconds

Average speed = $\frac{7}{5}$

Average speed =1.4 m/s

Average velocity = $\frac{Total \ displacement}{Total \ time}$

Total displacement = √4²+3²

Total displacement = √25

Total displacement = 5 metres

Total time = 5 seconds

Average velocity = $\frac{5}{5}$

Average velocity = 1 m/s