A particle under the action of a constant force moves from rest upto 20 seconds. if distance covered in first 10 seconds is s1 and then covered in nest 10 seconds is s2 then ? 1. s1=s2 2.s2=3s1 3. s2=2s1 4. s2=4s1

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Answer 1 · 2017-08-05T13:27:13

The answer is option $2$ $2$ , $s_{2} = 3 s_{1}$ $s_2=3s_1$

Let the acceleration be $= a$ $=a$ . It is a constant.

The initial velocity is $u = 0$ $u=0$

We apply the equations of motion for the first $10 s$ $10s$

$v = u + a t$ $v=u+at$

$v = 0 + 10 a$ $v=0+10a$

Then,

$v^{2} = u^{2} + 2 a s$ $v^2=u^2+2as$

$100 a^{2} = 0 + 2 a s_{1}$ $100a^2=0+2as_1$

$a = \frac{2 s_{1}}{100} = \frac{s_{1}}{50}$ $a=(2s_1)/(100)=s_1/50$

We apply the equations of motion for $20 s$ $20s$

$v = 0 + 20 a$ $v=0+20a$

$v^{2} = u^{2} + 2 a (s_{1} + s_{2})$ $v^2=u^2+2a(s_1+s_2)$

$400 a^{2} = 2 a (s_{1} + s_{2})$ $400a^2=2a(s_1+s_2)$

$s_{1} + s_{2} = 200 a = 200 \cdot \frac{s_{1}}{50} = 4 s_{1}$ $s_1+s_2=200a=200*s_1/50=4s_1$

$s_{2} = 3 s_{1}$ $s_2=3s_1$