Question

If the velocity of a body moving with uniform velocity change from 10m/s to20m/s in. 10s what is the acceleration of the body

Answer 1

GiveN :

• Initial velocity (u) = 10 m/s
• Final velocity (v) = 20 m/s
• Time interval (t) = 10 s

To FinD :

• Acceleration of the body

SolutioN :

Use 1st equation of motion :

$\large \implies {\boxed{\rm{v = u + at}}}$

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$\implies \rm{20 = 10 + 10 \times a}$

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$\implies \rm{20 - 10 = 10a}$

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$\implies \rm{10a = 10}$

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$\implies \rm{a = \dfrac{10}{10}}$

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$\implies \rm{a = 1}$

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$\small \underline {\rm{\therefore \: Acceleration \: of \: body \: is \: 1 \: ms^{-2}}}$

________________________

• Acceleration is defined as rate of change of velocity.

• SI unit of acceleration is m/s².

• Acceleration is vector quantity.

• $\small \sf{Acceleration \: = \: \dfrac{Final \: velocity \: - \: Initial \: velocity}{time}}$

Answer 2

GivEn:-

• Initial Velocity of a body (u) = $\sf 10ms^{-1}$

• Final Velocity of a body (v) = $\sf 20ms^{-1}$

• Time Interval or time taken to move a body (t) = 10s

To find:-

• Acceleration of the body

SoluTion:-

As we know the,

★ First equation of motion,

$\star\;{\underline{\boxed{\bf{\pink{v = u + at}}}}}$

$\;\;\;\sf\dag\; \underline{Put\;the\;givEn\;values\:-}$

$:\implies\sf 20 = 10 + 10 \times a \times a$

$:\implies\sf 20 - 10 = 10a$

$:\implies\sf 20 - 10 = 10a$

$:\implies\sf 10 = 10a$

$:\implies\sf a = \cancel{ \dfrac{10}{10}}$

$:\implies{\underline{\boxed{\bf{\pink{1\;ms^{-2}}}}}}$

$\therefore\sf \underline{Acceleration\;of\;body\;is\; \bf{1ms^{-2}}}$

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Additional Information:-

★ There are three equations of motion:-

⠀⠀⠀✩ $\sf v = u + at$

⠀⠀⠀✩ $\sf s = ut + \dfrac{1}{2} at^2$

⠀⠀⠀✩ $\sf v^2 - u^2 = 2as$

Where,

• v = Final velocity

• u = Initial velocity

• a = acceleration

• t = time

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$\;\;\star\;\sf Acceleration (a) = \dfrac{Final\; velocity (v) - Initial\; velocity (u)}{Time (t)}$

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