# A position function for rectilinear motion is given by x(t) = 1 / 3 t^3 - 1 / 2 t^2 + 6 t + 2 for...

## Question:

A position function for rectilinear motion is given by {eq}x(t) = \frac{1}{3} t^3 - \frac{1}{ 2 } t^2 + 6 t + 2 {/eq} for {eq}0 \le t \le 8. {/eq}
a) Find velocity {eq}x'(t), {/eq} and acceleration {eq}x''(t) {/eq} functions.
b) Find the interval when the object is moving left.
c) At what time(s) has the object stopped?
d) When will the acceleration be zero?
e) What will be the velocity when the acceleration is zero?

## Distance, Velocity, And Acceleration:

The indefinite integral is commonly applied in problems involving distance, velocity, and acceleration, each of which is a function of time.

The derivative of a distance function represents instantaneous velocity and that the derivative of the velocity function represents instantaneous acceleration at a particular time. In considering the relationship between the derivative and the indefinite integral as inverse operations, The indefinite integral of the acceleration function represents the velocity function and that the indefinite integral of the velocity represents the distance function.

Mathematically,

If {eq}s(t) {/eq} represents the position of the particle at any time {eq}t {/eq}, then

{eq}\eqalign{ & v\left( t \right) = \frac{{ds}}{{dt}} \cr & a\left( t \right) = \frac{{dv}}{{dt}} = \frac{{{d^2}s}}{{d{t^2}}} \cr} {/eq}

The position relation of the particle is given by,

{eq}x(t) = \dfrac{1}{3}{t^3} - \dfrac{1}{2}{t^2} + 6t + 2 {/eq}

The velocity is the derivative of position relation with respect to time.

Therefore,

{eq}v(t) = x'(t) = {t^2} - t + 6 {/eq}

When the object will be moving left, the velocity of the particle will be negative.

Since, this velocity is always positive,

{eq}\eqalign{ & [D < 0 \cr & a > 0] \cr & v(t) > 0 \cr} {/eq}

The acceleration of the object is the derivative of the velocity with respect to time.

Therefore, the acceleration is given by,

{eq}a(t) = x''(t) = 2t - 1 {/eq}

The instance, when the object stops, {eq}v(t) = 0 {/eq}

Therefore,

{eq}{t^2} - t + 6 = 0 {/eq}

{eq}t \notin R {/eq}

Since, no real number satisfy the above relation, the velocity throughout the motion is always strictly greater that zero.

The time when acceleration is zero is given by,

{eq}\eqalign{ & a = 0 \cr & 2t - 1 = 0 \cr & t = \dfrac{1}{2}s \cr} {/eq}

The velocity at this instance will be given by,

{eq}v\left( {\dfrac{1}{2}} \right) = {\left( {\dfrac{1}{2}} \right)^2} - \left( {\dfrac{1}{2}} \right) + 6 = 5.75unit/\sec {/eq}