Question

Usha swims in a 90 m long pool.she covers 180m in 1 min by swimming from 1 and2 the other and back a long the same Street path find the average speed and average velocity of usha

Answer 1

Hey mate
here is your answer

Length of the pool: 90m
Average speed:
Speed= Distance/Time
speed=180/60
speed=30 m/s

Velocity:
velocity=displacement/time

displacement=0( Since she comes back to the same place)

velocity=0

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 ,

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