Question

class 11 physics................. ​

Answer 1

Answer:

$\boxed{\mathfrak{Position \ of \ body \ at \ 1s = 0.5 \ m}}$

Explanation:

Relation between acceleration (a) of body w.r.t time (t) is given as:

a = 3t m/s²

Accelration is given as rate of change of velocity i.e.

$\rm \implies a = \dfrac{dv}{dt}$

So,

$\rm \implies \dfrac{dv}{dt} = 3t \\ \\ \rm \implies \int dv = \int (3t)dt \\ \\ \rm \implies v = \dfrac{ 3{t}^{2} }{2}$

Velocity is given as rate of change of position i.e.

$\rm \implies v = \dfrac{dx}{dt}$

So,

$\rm \implies \frac{dx}{dt} = \frac{3 {t}^{2} }{2} \\ \\ \rm \implies \int dx = \int (\dfrac{3 {t}^{2} }{2}) dt \\ \\ \rm \implies x = \dfrac{ {t}^{3} }{2}$

Position of body at t = 1s:

$\rm \implies x = \dfrac{ {1}^{3} }{2} \\ \\ \rm \implies x = \dfrac{1}{2} \\ \\ \rm \implies x = 0.5 \: m$