Physics, asked by roymr071, 6 months ago

a particle moving with a speed of 20m/s2 accelerated with 5m/s2.Find the distance in 4 seconds​

Answers

Answered by Asterinn
10

Given :

  • initial velocity = 20 m/s

  • acceleration = 5 m / s²

  • time taken = 4 seconds

To find :

  • Distance travelled

Formula used :

  • S = ut + 1/2 at²

where :-

  • s = Distance travelled
  • u = initial velocity
  • a = acceleration
  • t = time taken

Solution :

to find distance travelled we will use the following formula :-

⟹ S = ut + 1/2 at²

Now put :-

  • u = 20
  • a = 5
  • t = 4

⟹ S = (20×4) +[( 1/2) × (5) ×( 4²)]

⟹ S = (20×4) +[( 1/2) × (5) × 4 ×4]

⟹ S = 80 + (5× 2 ×4)

⟹ S = 80 + 40

⟹ S = 80 + 40

⟹ S = 120 m

Answer :

Distance travelled = 120 m

______________________________

Learn more :-

v= u+at

v= u+atv²-u²= 2as

v= u+atv²-u²= 2ass = ut+1/2at²

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Answered by ArcaneAssassin
115

 \huge\sf { \dag Question }

A particle moving with a speed of 20m/s^2 accelerated with 5m/s^2. Find the distance in 4 seconds.

 \huge\sf { \dag Solution }

 {\underline {\underline  {\rm {\red { Given }}}}}

  •  \sf\orange { u ( initial \: velocity ) = 20m/{s}^{2} }

  •  \sf\orange { t( total \: time  ) = 4s }

  •  \sf\orange { a( acceleration ) = 5m/{s}^{2} }

 {\underline {\underline  {\rm {\red { To \: find }}}}}

  •  \sf { s ( distance \: travelled ) }

 {\underline {\underline  {\rm {\red { Calculation }}}}}

According to the 2nd equation of motion-

 {\underline {\underline  {\rm {\red { \qquad s = ut + \dfrac{1}{2}a{t}^{2} }}}}}

Where,

  • s = distance travelled
  • u = intial velocity
  • t = time
  • a = acceleration

Substituting values-

 \sf { \qquad \leadsto s = 20 \times 4 +  \dfrac{1}{2} \times 5 \times 4 \times 4 }

 \sf { \qquad \leadsto s = 20 \times \dfrac{1}{2} \times 4 + 5 \times 4 \times 4 }

 \sf { \qquad \leadsto s = 80 + \dfrac{1}{ \cancel 2} \times 5 \times \cancel { {16}^{ \: \: 8 } }  }

 \sf { \qquad \leadsto s = 80 + 40 }

 \sf { \qquad \leadsto s = 120m }

Therefore,the distance in 4 seconds is 120m.

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 \huge\sf { \dag Additional \: Information }

THREE EQUATIONS OF MOTION-

  •  {\underline {\underline {\boxed {\rm {\green { v = u + at }}}}}}

  •  {\underline {\underline {\boxed {\rm {\green {  s = ut + \dfrac{1}{2}a{t}^{2} }}}}}}

  •  {\underline {\underline {\boxed {\rm {\green { 2as = {v}^{2} - { u}^{2} }}}}}}

To solve problems on motion we should always remember that-

  • If a body starts from rest or stationary position then u ( initial velocity ) = 0.

  • If a body comes to rest or breaks are applied to stop then, v ( final velocity ) = 0.

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