Question

plzz solve this equation​

Answer 1

Required Solution:

In this question we are using the third equation of motion to calculate the distance travelled, so:–

${\underline {\boxed {\Large {\sf { {v}^{2}= {u}^{2} + 2as} }}}}$

$\bf { ⇢ {0}^{2}= {20}^{2} + 2 \times (-8) \times s}$

$\bf { ⇢ 0 = 400 + 2 \times (-8) \times s}$

$\bf { ⇢ 0 = 400 + (-16) \times s}$

$\bf { ⇢ 0 = 400 + (-16s) }$

$\bf { ⇢ 0 = 400 - 16s }$

$\bf { ⇢ -16s = 0-400 }$

$\bf { ⇢ -16s = -400 }$

$\bf { ⇢ s = \dfrac{-400}{-16} }$

$\boxed { \bf { ⇢ s = 25m} }$

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More information!

Equations of motion:

• v = u + at
• s = ut + ½at²
• v² – u² = 2as

Where,

★ v = final velocity = m/s

★ u = initial velocity = m/s

★ a = acceleration = m/s²

★ s = distance/displacement = m

★ t = time = sec

Remember that!

• When a body starts from rest, its initial velocity is 0.

• When a body comes to stop or applies breaks, its final velocity is 0.

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