Question

A man goes once round of a rectangular field whose length and breadth is 4m and 3m in 20sec. Find the distance traveled and displacement after 2min and 10sec, also find average speed and average velocity
​

Answer 1

Distance

Perimeter of rectangular field = 2(l+b)

$= 2(4 + 3)$

$= 2(7)$

$= 14 \: m$

Distance travelled in 2min 10 sec

2 min 10 sec =130 sec

•

•

A man can travel 14m in 20 second

$20 \: sec = 14 \: m$

$1 \: sec = \frac{14}{20}$

Then in 130 sec :-

$130sec = \frac{14}{20} \times 130$

Distance travelled in 2 min 10 sec -

$\huge\red{= 91 \: m}$

•

•

•

•

•

Displacement

Suppose the man started running ABCD rectangular field from point A

$Displacement = \frac{91}{14}$

$Diplacement = 6.5$

It means the man completed 6 complete round and half round.

It means after starting from point A, he completed 6 rounds and now he is on point C

For displacement

Using Pythagoras formula

$AC {}^{2} = AB {}^{2} + BC {}^{2}$

$AC {}^{2} = (3) {}^{2} + (4) {}^{2}$

$AC {}^{2} = 9 + 16$

$AC{}^{2} = 25$

$AC = 5 \: m$

$Displacement = AC$

$\huge\red{\boxed{\green{Displacement = 5 \: m}}}$

•

•

•

•

•

Average Speed

$Speed = \frac{Distance}{Total \: Time}$

$Average \: Speed = \frac{91}{130}$

$\huge\red{\boxed{\green{Average \: Speed = 0.7 \: ms {}^{ - 1}}}}$

•

•

•

•

•

Average Velocity

$Average \: Velocity = \frac{Change \: in \: Displacement}{Change \: in \: Time}$

$Average \: Veloity = \frac{5}{130}$

$\huge\red{\boxed{\green{Average \: Velocity = 0.038 \: ms {}^{ - 1}}}}$

•

•

•

•

•

MARK BRAINLIEST