Question

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

Answer 1

Answer:

Average speed = 3 ms⁻¹

Average velocity = 0 ms⁻¹

Explanation:

Total distance covered by usha in 1 minute = 180 metres

Displacement of Usha in 1 minute = 0 metres

Average speed = Total distance covered / total time taken

= 180/1 min

1 min = 60 seconds

180/60 = 3 m s⁻¹

Average velocity = Displacement / time taken

0/60 seconds

= 0ms⁻¹

The average speed of Usha is 3 ms⁻¹ and average velocity is 0 ms⁻¹

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 ,

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