Question

The distance travelled by a body as a function of time is given by X = t^3 - 3t^2 + 6 where X is M and T is in seconds find the minimum and maximum displacement of the body from the origin

Answer 1

Answer:

The distance travelled by a body as a function of time is given by X = t^3 - 3t^2 + 6 where X is M and T is in seconds find the minimum and maximum displacement of the body from the origin

Answer 2

Answer:

The maximum displacement ant the minimum displacement of the body from origin is 6 m and 2 m respectively.

Explanation:

Distance travelled by a body is

$x(t)=t^3-3t^2+6$ ...... (1)

Differentiate the equation (1) with respect to $t$ as follows:

$\frac{dx}{dt}=\frac{d}{dt}( t^3-3t^2+6)$

    $=3t^2-6t$ ...... (2)

Compute the critical points by letting $\frac{dx}{dt}=0$.

⇒ $3t^2-6t=0$

⇒ $3t(t-2)=0$

⇒ $3t=0$ and $t-2=0$

⇒ $t=0$, $t=2$

Differentiate the equation (2) with respect to $t$ as follows:

$\frac{d^2x}{dt^2}=\frac{d}{dt}( 3t^2-6t)$

     $=6t-6$

     $=6(t-1)$ ...... (3)

If $\frac{d^2x}{dt^2} < 0$ corresponding to $t$, then $x$ will be maximum.

If $\frac{d^2x}{dt^2} > 0$ corresponding to $t$, then $x$ will be minimum.

Substitute the value $0$ for $t$ in (3), we get

$\frac{d^2x}{dt^2} = 6(0-1)$

     $=-6$

This implies, $\frac{d^2x}{dt^2} < 0$ for $t=0$.

Thus, maximum displacement of the body from origin = $(0)^3-3(0)^2+6$

                                                                                            = $6$ m

Similarly,

Substitute the value $2$ for $t$ in (3), we get

$\frac{d^2x}{dt^2} = 6(2-1)$

     $=6$

This implies, $\frac{d^2x}{dt^2} > 0$ for $t=2$.

Thus, minimum displacement of the body from origin = $2^3-3(2)^2+6$

                                                                                           = $8-12+6$

                                                                                           = $2$ m

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