Question

usha swims in 90m long pool.she covers 180m in 1 min by swimming from one end to other back along the same straight path find
1) Average speed
2)Average velocity
useless answers would be reported.,..
please guys solve it​

Answer 1

Given :

It is given that :

• Length ofthe pool = $90m$
• Length covered in $1~min$ = $180m$ i.e., A to B

To find :

We've to find the following :

• Average speed
• Average velocity

Solution :

Firstly, we'll find the average speed :

• Total distance = $180m$
• Total time = $60~secs$

$:\implies Average ~speed = \dfrac{Total~distance}{Total~time}$

$:\implies Average~ speed = \dfrac{180m}{60~secs}$

$\red{\therefore ~\bold{Average ~speed=3\dfrac{m}{s}}}$

Now, we'll find the average velocity :

• Displacement between A and B = 0
• Time taken = $60~secs$

$:\implies Average ~velocity=\dfrac{Displacement}{Time~taken}$

$:\implies Average ~velocity=\dfrac{0}{60~secs}$

$\red{\therefore~\bold{Average~velocity=0\dfrac{m}{s}}}$

Answer 2

Answer:

Average speed = 3 m/s

Average velocity = 0 m/s

Explanation:

Given :-

Usha swims in a 90 m long pool.

She covers 180 m in 1 minute by swimming from one end to other back along the same straight path.

To find :-

Usha's average speed and average velocity.

Solution :-

We have,

Distance = 180 m

Time = 1 minute = 60 seconds

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

• $\frac{180}{60}$

• $3$

∴ Average speed = 3 m/s

As Usha swims and back to the same path where she started, hence the displacement will be zero as the initial and final position are the same.

Average velocity = $\frac{Displacement}{Time}$

• $\frac{0}{60}$

• $0$

∴ Average velocity = 0 m/s