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

Answer:

The velocity is - 5.5m/s

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

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

$v(0.5) = -5.5$