Question

A body moves in a straight line with uniform acceleration of 2 m/s2. If the initial
velocity of the body is 16 m/s, find its final velocity after covering 36 m of
distance using the third equation of motion. *
O a) v = 10 m/s
O b) v = 15 m/s
O c) v = 6 m/s
O d) v = 20 m/s​

Answer 1

Answer.

Given -

$\longrightarrow$a = 2 m/s²

$\longrightarrow$u = 16 m/s

$\longrightarrow$s = 36m

where a is acceleration , u is initial velocity and s is distance.

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To find -

$\longrightarrow$Final velocity ( v )

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Formula used -

$\longrightarrow$v² = u² + 2as

where a is acceleration , u is initial velocity , v is the final velocity and s is distance.

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Solution -

$\longrightarrow$${v}^{2} = {16}^{2} + 2 \times 2 \times 36$

$\longrightarrow$${v}^{2} = 256 + 144$

$\longrightarrow$${ v }^{2} = 400$

$\longrightarrow$$v = \sqrt{400}$

$\longrightarrow$$v = 20 m/s$

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ADDITIONAL INFORMATION -

$\implies$v = u + at

$\implies$v² = u² + 2as

$\implies$s = ut + ½at²

$\implies$Snth = u + (2n - 1)a/2

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Thanks

Answer 2

$\huge\underline{\pink{A}\orange{N}\blue{S}\purple{W}\red{E}\orange{R}}$

Given -

⟶ a = 2 m/s²

⟶ u = 16 m/s

⟶ s = 36m

where a is acceleration , u is initial velocity and s is distance.

━━━━━━━━━━━━━━━━━━━━━━━━━━

To find -

⟶ Final velocity ( v )

━━━━━━━━━━━━━━━━━━━━━━━━━━

Formula used -

⟶ v² = u² + 2as

where a is acceleration , u is initial velocity , v is the final velocity and s is distance.

━━━━━━━━━━━━━━━━━━━━━━━━━━

Solution -

$⟶ {v}^{2} = {16}^{2} + 2 \times 2 \times 36$

$⟶ {v}^{2} = 256 + 144$

$⟶ { v }^{2} = 400v$

$⟶ v = \sqrt{400}$

$⟶ v = 20 m/s$

━━━━━━━━━━━━━━━━━━━━━━━━━━

ADDITIONAL INFORMATION -

⟹ v = u + at

⟹ v² = u² + 2as

⟹ s = ut + ½at²

⟹ Snth = u + (2n - 1)a/2

━━━━━━━━━━━━━━━━━━━━━━━━━