Question

Usha swims in a 90m long pool.she covers 180m in 1minute 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

Explanation:

Average speed = Total distance/ time

= 180/ 1/60

= 10800 m/h

= 10.8 km/h

Average velocity = displacement/time

= 0/ 1/60

= 0 km/h

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 ,

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