Question

the position x of a particle varies with time t as x=6+12t-2t^2
What is - 1)instant velocity
2)instant acceleration ​

Answer 1

Given that,

At some position x,x=6+12t-2t²

To find:

Instantaneous Velocity and Instantaneous Acceleration of the particle

Using the formula,

$\sf{ \frac{dy}{dx} = nx {}^{n - 1} }$

We know that,

Instantaneous Velocity= dx/dt

→v= d[6+12t-2t²]/dt

→v=0+12-4t

→v=12-4t

Thus,the velocity of the particle at some time t is (12-4t)m/s

Now,

Instantaneous Acceleration=dv/dt

»a=dv/dt

»a=d[12-4t]/dt

»a=0-4m/s²

»a= -4m/s²

Here,the particle shows retardation