Question

a particle moves along the x axis such that position x with respect to origin is given by x=bt^2 where b=0.6ms^-2.its accelaration during t=1s to t=5s is

Answer 1

Given:

A particle moves along the x axis such that position x with respect to origin is given by x=bt^2 where b=0.6ms^-2.

To find:

Acceleration of the particle between t = 1 sec to t = 5 sec.

Calculation:

Acceleration can be calculated by double differentiation of the displacement function with respect to time.

$\therefore \: x = b {t }^{2}$

Differentiating Displacement w.r.t time:

$= > v = \dfrac{dx}{dt}$

$= > v = \dfrac{d(b {t}^{2}) }{dt}$

$= > v = b \times \dfrac{d( {t}^{2}) }{dt}$

$= > v = 2bt$

Again differentiating velocity w.r.t time:

$= > a = \dfrac{dv}{dt} = \dfrac{ {d}^{2} x}{d {t}^{2} }$

$= > a = \dfrac{d(2bt)}{dt}$

$= > a = 2b \times \dfrac{d(t)}{dt}$

$= > a = 2b \times 1$

$= > a = 2b$

$= > a = 2 \times (0.6)$

$= > a = 1.2 \: m {s}^{ - 2}$

So , acceleration is constant for the particle and is equal to 1.2m/s² inbetween the given time interval.