Question

Usha swims in a 19 long pool she covers 180 metre in one minute by swimming from one end to another and back long the same path find the average speed and average velocity of Usha

Answer 1

HEY BUDDY !!!!

HERE IS YOUR ANSWER ⤵️⤵️⤵️

THE POOL WAS LONG = 90M

SHE COVERS DISTANCE IN ONE MINUTE =180M

AVERAGE SPEED

= TOTAL DISTANCE COVERED / TOTAL TIME TAKEN

1 MINUTE = 60 SECONDS

= 180/60

= 3m/s

AVERAGE VELOCITY

= DISPLACEMENT /TIME

=0/60

=0m/s

SO THE AVERAGE SPEED IS 3m/s AND THE AVERAGE VELOCITY IS 0m/s

HOPE THIS HELPED...

MARK IT AS BRAINLIEST...

AND PLEASE FOLLOW ME...

AND I HAD ALSO ASKED THIS QUESTION...

Answer 2

Answer:-

Explanation :-

Usha Covers a distance (d)= 180 m at a

time 1 minute or 60 seconds .

Usha return to initial position ( displacement ) = 0

We know that,

$\bf{average \: speed \: = \frac{total \: distance}{total \: time \: } } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{180}{60} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: 3 \: \frac{m}{s}$

And ,

$\bf{average \: velocity \: = \frac{displacement}{time} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{0}{60} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 0 \: \: \frac{m}{s}$