Question

A body start from origin and moves along x-axis so that it's position at any instant is x =4t^2-12t where t is in second and v is m/s.what is acceleration of partial ?

Answer 1

$x = 4 {t}^{2} - 12t \\ \frac{dx}{dt } = 8t - 12 = v \\ a = \frac{dv}{dt} = 8$
hence the acceleration is
$8 \frac{m}{ {s}^{2} }$

Answer 2

Answer:

The acceleration is 8 m/s².

Explanation:

Given that,

Position $x = 4t^2-12t$....(I)

Where, t in sec and v in m/s

The velocity is the first derivative of the position of the object.

On differentiating equation (I) w.r.to t

$v=\dfrac{dx}{dt}=8t-12$....(II)

The acceleration is the rate of change of velocity of the object or the second derivative of the position of the object.

On differentiating equation (II)

The acceleration is

$a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}=8$

Hence, The acceleration is 8 m/s².