Question

A particle start it motion from rest under the action of constant force if the distance covered in first 10 second is si ​

Answer 1

Answer:

if the particle start motion see the action of constant force distance is 10 you can - from 20=10, so the answer is 10

Answer 2

Explanation:

$\huge\bf\underline\blue{Solution\::-}$

If the particle is moving in a straight line under the action of a constant force or under constant acceleration (a) Using,

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

Since the body starts from the rest

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: u = 0$

since,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: s = \frac{1}{2} a {t}^{2}$

Now,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: s_{1} = \frac{1}{2} s( {10})^{2}$

..... ( i ) and

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: s_{2} = \frac{1}{2} a( {20})^{2}$

.....( ii )

Dividing Eq. (i) and Eq. (ii), we get

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{ s_{1}}{ s_{2} } = \frac{( {10})^{2} }{( {20})^{2} }$

$= > s_{2} = 2 s_{1}$

PLEASE FOLLOW ME....