Usha swims in 90m long pool. She covers 180 m in 1 min by swimming from one end to the other and back along the same straight path. Find the average speed and average velocity of Usha.
Answers
Answer:
average speed=3m/s
average velocity=0
Explanation:
length of pool = 90m
total distance covered by Usha = 180m
time taken = 1min=60s
average speed = 180/60=3m/s
since displacement will be zero since she returns to the starting point
therefore average velocity will be 0
Explanation:
Total distance covered by usha in 1 min is 180 m.
Displacement if usha in 1 min=0 m
\begin{gathered} \bf \: Average \: speed = \frac{Total \: Distance \: covered }{Total \: time \:taken } \\ \end{gathered}
Averagespeed=
Totaltimetaken
TotalDistancecovered
\begin{gathered} \sf = \frac{180 \: m}{1 \: min} = \frac{180m}{1min} \times \frac{1min}{60 \: s} \\ \\ \\ \boxed{ \bf \: = 3 {m}^{s - 1}} \end{gathered}
=
1min
180m
=
1min
180m
×
60s
1min
=3m
s−1
\begin{gathered} \bf \: Average \: speed = \frac{Displacement}{Total \: time \: taken} \\ \\ \\ \\ \\ = \sf\frac{0m}{60s} \\ \\ \\ \\ \\ = \boxed{ \sf \: 0 {m}^{s - 1}} \end{gathered}
Averagespeed=
Totaltimetaken
Displacement
=
60s
0m
=
0m
s−1
The average speed of Usha is 3m^s-1 and her average velocity is 0m^s-1.