Question

A particle moves along x axis and its displacement at any time is given by x(t)=2t^3-3t^2-4t in SI Units.The velocity of the particle when its acceleration is zero is?​

Answer 1

The velocity is - 5.5m/s

Step-by-step explanation:

Given

`x(t) = 2t^3 - 3t^2 - 4t`

Required

The velocity when acceleration = 0

First, calculate the velocity (v)

`v = x'(t)`

`x(t) = 2t^3 - 3t^2 - 4t` --- differentiate

`x'(t) =6t^2 - 6t - 4`

So:

`v(t) =6t^2 - 6t - 4`

Next, calculate acceleration (a)

`a = v'(t)`

`v(t) =6t^2 - 6t - 4` --- differentiate

`v'(t) = 12t - 6`

So:

`a(t) = 12t - 6`

From the question acceleration = 0

So:

`a(t) = 12t - 6 =0`

`12t - 6 =0`

Solve for t

`12t = 6`

`t = 6/12`

`t=0.5`

Substitute
`t=0.5` in
`v(t) =6t^2 - 6t - 4` to get the velocity

`v(0.5) = 6 * 0.5^2 - 6 * 0.5 - 4`

`v(0.5) = -5.5`