Question

Usha swims in a 90 m long
pool. She covers 180 m in one minut
by swimming from one end to the other
and back along the same straight path
Find the average speed and average
velocity of Usha.

Answer 1

Answer:

average speed = total distance/ total time

note: time must be in seconds

180/60 = 3m/s

average velocity = total displacements/ total time

displacement is zero so the average velocity is zero

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 ,

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