Question

A body starts from rest with a uniform acceleration of 2 meter per second square. Find the distance covered by the body in 2 seconds .​

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

$\tt Given \begin{cases} \sf{Initial \: Velocity \: (u) \: = \: 0 \: ms^{-1}} \\ \sf{Accelaration \: (a) \: = \: 2 \: ms^{-2}} \\ \sf{Time \: interval \: (T) \: = \: 2 \: s} \end{cases}$

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To Find :

Distance travelled by body

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Solution :

We have formula :

$\Large{\underline{\boxed{\sf{S \: = \: ut \: + \: \frac{1}{2} \: at^2}}}}$

Where,

u is initial velocity

t is time interval

a is acceleration

S is distance travelled by body

Put Values

⇒S = 0(2) + ½ (2)(2)²

⇒S = 0 + ½(2)(4)

⇒S = ½(8)

⇒S = 4

$\Large{\underline{\boxed{\sf{Distance \: Travelled \: = \: 4 \: m}}}}$

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All Equation of Motion are :

• v = u + at

• S = ut + 1/2 at^2

• v^2 - u^2 = 2as

Where v is final velocity

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#BAL

Answer 2

$\Huge{\underline{\underline{\sf{Answer \colon}}}}$

From the Question,

• Initial Velocity (u) = 0 m/s

• Acceleration (a) = 2 m/s²

• Time (t} = 2 s

To find

Distance covered by the object in 2s

Using the Relation,

$\huge{ \boxed{ \boxed{ \sf{s = ut + \frac{a {t}^{2} }{2} }}}}$

Putting the values,we get :

$\large{ \implies \: \sf{s = (0)(2) + \frac{2 \times 2 {}^{2} }{2} }} \\ \\ \large{ \implies \: \sf{s = \frac{8}{2} }} \\ \\ \huge{ \implies \: \boxed{ \boxed{ \sf{s = 4m}}}}$

The object covers 4m in 2s after starting from rest

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