Question

X = 3 t^2+ 5 t + 12 then the initial position of the particle is?​

Answer 1

Answer:

The displacement is given as x=5t

2

+4t+3

⇒x−3=5t

2

+4t

assuming x−3=s,

s=4t+

2

1

10t

2

comparing this with s=ut+

2

1

at

2

we get u=4m/s and a=10m/s

2

so velocity at t=2sec is v=u+at=4+10×2=24m/s

Answer 2

$\color{red}\huge{\underline{\underline{GIVEN\dag}}}$

• Displacement as a function of time.
• $X = 3 {t}^{2} + 5t + 12$
• For initial position $t=0$

$\color{red}\huge{\underline{\underline{SOLUTION\dag}}}$

$X = 3 {(0)}^{2} + 5(0) + 12$

$\orange{ \boxed{\therefore X = 12}}$

Initially the paricle was at 12 units.

HOPE THAT HELPS!!