Question

A particle after starting from rest experiences constant acceleration for 20 seconds if it covers a distance of S1 in first 10 seconds and distance S2 in next 10 second then tell relation S2 and S1.

Answer 1

starting from rest means
initial velocity (u)=0

●acceleration is constant so we can use equation of motion

$S1= ut + \frac{1}{2} a {t}^{2} \\ = 0 + \frac{1}{2} a \: 100 \\ = 50a \:$
S1=50a

Now come to the next part-

Let total distance it covered is S
means

S=S1+S2

$s = ut + \frac{1}{2} a { \: t}^{2} \\ 0 + \frac{1}{2} a \: {20}^{2} \\ = 200a$

S=200a
S1=50a

S2=S-S1
S2=200a-50a

S2=150a


S1=50a
S2=150a

◆◆Relation is S2=3S1◆◆

Hope it will help you.

Answer 2

ANSWERS :-

$solution$

Apply equation of motion from O to A.
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S1 = ut + 1/2 at²

S1 = 0 * t + 1/2 at²

S1 = 1/2at² .........(1)

apply equation of motion from O to B
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S = ut + 1/2 at²

S1 + S2 = 0(2t) + 1/2 a(2t)²

S1 + S2 = 0 + 1/2 a 4t²

S1 + S2 = 2at²

1/2 at² + S2 = 2at²

S2 = 2at² - 1/2 at²

S2 = 3/2 at² ........(ii)

From equation (I) and (ii)

S2/S1 = 3/2 at² / 1/2at²

S2/S1 = 3

✨ Relation S2 = 3S1

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❤BE BRAINLY❤
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