Question

pls answer me this question fast....​

Answer 1

AnsWer :

2 seconds.

SolutioN :

Let,

• acceleration ( a )
• u initial
• v final
• m mass

→ Force / mass = 20 - 10 / 2 + 1 = 10 / 3.

$\tt \: F_{net}\: = m \times a.$

$\tt : \implies a = \frac{F_{net}}{mass}$

$\tt : \implies a = \frac{10}{2 + 1}$

$\tt : \implies a = \frac{10}{3} \: m / {s}^{2} .$

Now,

$\tt S = ut + \dfrac{1}{2} \times a { t}^{2}$

Where as,

• S = 20/3
• u = 0.
• a = 10/3.
• t = ?

$\tt : \implies\dfrac{20}{3} = 0 \times t + \dfrac{1}{2} \times \dfrac{10}{3} \times { t}^{2}$

$\tt : \implies\dfrac{20}{3} = 0 \times t + \dfrac{1}{ \cancel2} \times \dfrac{ \cancel{10}}{3} \times { t}^{2}$

$\tt : \implies\dfrac{ \cancel{20}}{ \cancel3} = \dfrac{ \cancel5}{ \cancel3} \times { t}^{2}$

$\tt : \implies {t}^{2} = 4.$

$\tt : \implies t = \pm2.$

★ Negative ignore,

→ t = 2.

Therefore, time taken to become distance between 0 is 2 seconds.