Question

Usha swims in a 90m long pool. She covers 180m in one minute 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

$\huge\bigstar\underline\mathfrak{Answer:-}$

Given:-

Total distance covered by Usha in 1 min is 180m.

Displacement of Usha in 1 min = 0m

$average \: speed = \frac{total \: distance \: covered}{total \: time \: taken}$

$\frac{180 \: m}{1 \: min} = \frac{180 \: m}{1 \: min} \times \frac{1 \: min}{60 \: s}$

$= 3m/s$

$average \: velocity = \frac{displacemet}{total \: time \: taken}$

$\frac{0 \: m}{60 \: s}$

$= 0m/s$

Answer 2

$\small \bold \pink{ \underline{ \underline{Answer }}}$

Given

• Total distance covered by Usha in 1 min is 180m.

• Displacement of Usha in 1 min = 0m

Average speed = Total distance/total time

➛ 180m/1min

⛬ 180m/60s

So, average speed = 3m/s

Average velocity = Displacement/Total Time

➛$\boxed{ \sf{ \frac{0m}{60s} } }$

➛ 0 m/s