Question

The position of a particle is given by y=kt^3 where k is constant with the dimmensions of m/s^3. Calculate the acceleration of a particle as a function of a timr

Answer 1

$y = k \times {t}^{3}$
$v = \frac{dy}{dt} = 3 \times k {t}^{2} \\ a = \frac{dv}{dt} = 6 \times k \times t$

Answer 2

Answer:

$a=6kt$

Explanation:

The position of a particle is given by :

$y=kt^3$...........(1)

Where

k is a constant and having the dimensions of $m/s^3$

Differentiating equation (1) to get velocity as :

$v=\dfrac{dy}{dt}$

$v=\dfrac{d(kt^3)}{dt}$

$v=3kt^2$...............(2)

Differentiating equation (2) to get acceleration as :

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

$a=\dfrac{d(3kt^2)}{dt}$

$a=6kt$

So, the acceleration of the particle is 6kt. hence, this is the required solution.