Question

if a particle moving in straight line such that its position varies with time as x=5(t-2) + 6(t-2)^2 ,then intial acceleration is​

Answer 1

Answer:The initial acceleration is $12 m/s^2$

Explanation:

Given that,

Position is

$x = 5(t-2)+6(t-2)^2$

$x = 6t^2+14-7t$.......(I)

On Differentiating

The velocity is

$v = \dfrac{dx}{dt}= 12t-7$

On differentiating

The acceleration is

$a = \dfrac{dv}{dt}= 12$

Hence, The initial acceleration is $12 m/s^2$