Physics, asked by prau, 1 month ago

A swimmer swims 90m long pool. He covers the distance of 180m by swimming from
one end to other end back along the same path. If he covers the first 90m at speed of 2m/s,
then how fast he swim so that his average speed is 3m/s?

Answers

Answered by FloralSparks
0

{\rm{\blue{\underline{\underline{\huge{❥Question}}}}}} \huge{\pink{࿐}}

A swimmer swims 90m long pool. He covers the distance of 180m by swimming from

one end to other end back along the same path. If he covers the first 90m at speed of 2m/s,

then how fast he swim so that his average speed is 3m/s?

\huge{ \pink{\mathfrak{ \underline{\overline{\mid \:\: ❥answer ✮\:\: \mid}}}}}

\huge{ \green{\mathfrak{ \underline{\overline{\mid \:\: ❥☆▪6m/s☆✮\:\: \mid}}}}}

{\rm{\red{\underline{\underline{\huge{❥Solution}}}}}} \huge{\pink{࿐}}

Let ,the time taken by the swimmer to swim from one end to other= t1

Distance from one end to other= 90m

And speed= 2m/s

{t1= 90/2}

⠀ =45 sec.

Let, the remaining time be t2

Remaining distance= 90m

And remaining time= t2

Total distance= 180m

Total time= 180s + t2

⠀⠀⠀⠀⠀⠀⠀\large{ \green{\mathfrak{ \underline{\overline{\mid \:\: ❥45+t2= 180/3\:\: \mid}}}}}

⠀⠀⠀⠀\large{ \green{\mathfrak{ \underline{\overline{\mid \:\: ❥180/45+t2= 3m/s\:\: \mid}}}}}⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀\large{ \green{\mathfrak{ \underline{\overline{\mid \:\: ❥45=t2= 60\:\: \mid}}}}}⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀\large{ \green{\mathfrak{ \underline{\overline{\mid \:\: ❥t2= 60 - 45\:\: \mid}}}}}⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀\large{ \green{\mathfrak{ \underline{\overline{\mid \:\: ❥= 15 sec\:\: \mid}}}}}⠀⠀⠀⠀

\huge{ \pink{\mathfrak{ \underline{\overline{\mid \:\: Speed required= 90/15\:\: \mid}}}}}

⠀⠀⠀⠀⠀\huge{ \pink{\mathfrak{ \underline{\overline{\mid \:\: ❥= 6m/s✮\:\: \mid}}}}}

⠀⠀⠀⠀⠀

Answered by Anonymous
0

Answer:

Let ,the time taken by the swimmer to swim from one end to other= t1

Distance from one end to other= 90m

And speed= 2m/s

{t1= 90/2}t1=90/2

⠀ =45 sec.

Let, the remaining time be t2

Remaining distance= 90m

And remaining time= t2

Total distance= 180m

Total time= 180s + t2

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