Question

A particle of mass 2 kg moving with a constant
acceleration covers a distance of 10 m in the
third second and 16 m in the 4th second. The
initial velocity of the particle is
-5 m/s (b) 6 m/s
(c) 10 m/s
(d) 8 m/s.

Answer 1

$\huge\underline{\underline{\bf \green{Question-}}}$

A particle of mass 2 kg moving with a constant acceleration covers a distance of 10 m in then third second and 16 m in the 4th second. The initial velocity of the particle is

$\huge\underline{\underline{\bf \green{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Mass of Particle ( m ) = 2 kg
• ${\sf s_1=10m\:\:\:t_1=3^{rd}s}$
• ${\sf s_2=16m\:\:\:t_2=4^{th}}$

$\large\underline{\underline{\sf To\:Find:}}$

• Initial velocity of Particle ( u )

✪$\large{\boxed{\bf \blue{s=u+\dfrac{a}{2}(2n-1) }}}$

$\implies{\sf s_1=u+\dfrac{a}{2}{+2n-1)}}$

$\implies{\sf 10=u+\dfrac{a}{2}(2×3-1)}$

$\implies{\sf 10=u+\dfrac{5a}{2}}$

$\implies{\sf \orange{20=2u+5a}\:\:\:\:\:→(1)}$

$\implies{\sf s_2=u+\dfrac{a}{2}(2n-1) }$

$\implies{\sf 16=u+\dfrac{a}{2}(2×4-1)}$

$\implies{\sf 16=u+\dfrac{7a}{2} }$

$\implies{\sf \orange{32=2u+7a}\:\:\:\:\:→(2)}$

❏ Subtracting eq 1 by eq 2

$\implies{\sf 20-32=(u+5a)-(u+7a) }$

$\implies{\sf -12=u+5a-u-7a }$

$\implies{\sf -12=-2a}$

$\implies{\sf \red{a=6\:m/s^2} }$

❏ Putting Value of 'a' in eq 1

$\implies{\sf 20=<u>2</u>u+5a}$

$\implies{\sf 20=2u+5×6}$

$\implies{\sf 20=2u+30}$

$\implies{\sf 2u=20-30 }$

$\implies{\sf u=\dfrac{-10}{2}}$

$\implies{\bf \red{u=-5\:m/s} }$

$\huge\underline{\underline{\bf \green{Answer-}}}$

Initial velocity ${\bf \red{(u)=-5\:m/s}}$