Question

The position of a particle moving along an x axis is given byx=12t²-2t³, where x is in meters and t is inseconds. Determine (a) the position, (b) the velocity, and (c) theacceleration of the particle at t=3.0s. (d) What is the maximumpositive coordinate reached by the particle and (e) at what time isit reached? (h) What is the acceleration of the particle at theinstant the particle is not moving (other than at t=0)? (i)Determine the average velocity of the particle between t=0 andt=3s.

Answer 1

Concepts and reason

The concepts required to solve the given problem are velocity, acceleration, average value of a function, and extreme points of a polynomial.

Evaluate the position at given value of time, $t = 3{\rm{ s}}$ , next use the definition of velocity and acceleration to evaluate the velocity and acceleration at the desired time. Further, use differential calculus to find the positive maxima of the given function of particle’s position, including the time when the particle attains that position.

Furthermore, equate the expression of velocity to zero to determine the time when the particle comes to rest and then evaluate the expression of acceleration at this time.

Finally, use the formula for average of a function to determine the average velocity of the particle within the given time interval.

Fundamentals

Velocity: It is a vector quantity defined as the rate of change of distance along a particular direction with respect to time; mathematically, expressed as,

$v = \frac{{ds}}{{dt}}$

Here, $v$ is velocity of the object, $s$ is the displacement of the object, and $t$ is time.

Acceleration: It is a vector quantity defined as the rate of change of velocity along a particular direction with respect to time; mathematically, expressed as,

$a = \frac{{dv}}{{dt}}$

Here, $a$ is acceleration of the object.

Extreme points of a polynomial are evaluated as,

Step 1. Determine the critical points of the function by evaluating the first derivative of the function with respect to the independent parameter and then equate the resulting equation to zero to solve for the critical points.

Step 2. Evaluate the second derivative of the provided function and evaluate the values of the resulting equation at the critical points to check the sign of the solution.

If the solution is less than zero then there’s a maximum at the critical point; if the solution is greater than zero there’s a minimum at the critical point.

However, if the solution is equivalent to zero then for values less than the critical point if the function is increasing and for the values greater than the critical point if the function is decreasing, then there’s a maximum at the critical point. While, for values less than the critical point if the function is decreasing and for the values greater than the critical point if the function is increasing, then there’s a minimum at the critical point.

Average value of a function $f\left( x \right)$ , in an interval $\left[ {a,b} \right]$ is evaluated as,

${F_{avg}} = \left( {\frac{1}{{b - a}}} \right)\int_a^b {f\left( x \right)dx}$

Here, ${F_{avg}}$ is the average value of the function.

(a)

Rewrite the given expression of the particle’s position at time t.

$x = 12{t^2} - 2{t^3}$

Substitute $3.0$ for t.

$\begin{array}{c}\\x = 12{\left( {3.0} \right)^2} - 2{\left( {3.0} \right)^3}\\\\ = 54{\rm{ m}}\\\end{array}$

(b)

Rewrite the given expression of the particle’s position at time t.

$x = 12{t^2} - 2{t^3}$

Evaluate the first derivative of the above equation.

$\begin{array}{c}\\v = \frac{{dx}}{{dt}}\\\\v = \frac{d}{{dt}}\left( {12{t^2} - 2{t^3}} \right)\\\end{array}$

$v = 24t - 6{t^2}$ …… (1)

Substitute $3.0$ for t.

$\begin{array}{c}\\v = 24\left( {3.0} \right) - 6{\left( {3.0} \right)^2}\\\\ = 18{\rm{ m/s}}\\\end{array}$

(c)

Rewrite the expression of the particle’s position at time t.

$x = 12{t^2} - 2{t^3}$

Evaluate the second derivative of the above equation.

$\begin{array}{c}\\a = \frac{{{d^2}x}}{{d{t^2}}}\\\\a = \frac{{{d^2}}}{{d{t^2}}}\left( {12{t^2} - 2{t^3}} \right)\\\end{array}$

$a = 24 - 12t$ …… (2)

Substitute $3.0$ for t.

$\begin{array}{c}\\a = 24 - 12\left( {3.0} \right)\\\\ = - 12{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

(d)

The first derivative of the function for particle’s position is equivalent to velocity. Thus, equate the velocity (equation 1) to zero to solve for the critical points as,

$\begin{array}{l}\\v = 0\\\\24t - 6{t^2} = 0\\\end{array}$

Thus,

Further, evaluate the second derivative of the equation for particle’s position to check for maximum.

Since, the second derivative for particle’s position is equivalent to acceleration. Thus, evaluate the expression of acceleration (equation 2) at critical point as,

$\begin{array}{c}\\{a_{t = 4{\rm{ s}}}} = 24 - 12\left( 4 \right)\\\\ = - 24{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Since, ${a_{t = 4{\rm{ s}}}} < 0$ , therefore,

At $t = 4{\rm{ s}}$ , the particle attains maximum position ${x_{\max }}$ , along positive x axis.

Substitute $t = 4{\rm{ s}}$ in the given expression of particle’s velocity.

$\begin{array}{c}\\{x_{\max }} = 12{\left( 4 \right)^2} - 2{\left( 4 \right)^3}\\\\ = 64{\rm{ m}}\\\end{array}$

(e)

As calculated in step 4, the particle attains maximum position at the critical point, that is, $t = 4.0{\rm{ s}}$ .

(h)

Rewrite the expression of velocity.

$v = 24t - 6{t^2}$

Equate the velocity to zero to solve for t,

$24t - 6{t^2} = 0$

Thus,

Rewrite the expression of acceleration.

$a = 24 - 12t$

Substitute $4.0$ for t.

$\begin{array}{c}\\a = 24 - 12\left( 4 \right)\\\\ = - 24{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

(i)

Rewrite the expression of velocity.

$v = 24t - 6{t^2}$

Calculate average velocity from $0{\rm{ s}}$ to ${\rm{3}}{\rm{.0 s}}$ as,

${v_{avg}} = \left( {\frac{1}{{{t_2} - {t_1}}}} \right)\int_{{t_1}}^{{t_2}} {vdt}$

Here, ${t_1}$ and ${t_2}$ are initial and the final time, respectively, and ${v_{avg}}$ is the average velocity.

Substitute $3.0{\rm{ s}}$ for ${t_2}$ , $0{\rm{ s}}$ for ${t_1}$ , and $24t - 6{t^2}$ for $v$ .

$\begin{array}{c}\\{v_{avg}} = \left( {\frac{1}{{3 - 0}}} \right)\int_0^3 {\left( {24t - 6{t^2}} \right)dt} \\\\ = \frac{1}{3}\left[ {\int_0^3 {\left( {24t} \right)dt} - \int_0^3 {\left( {6{t^2}} \right)dt} } \right]\\\\ = \frac{1}{3}\left[ {24\left( {\frac{{{t^2}}}{2}} \right)_0^3 - 6\left( {\frac{{{t^3}}}{3}} \right)_0^3} \right]\\\\ = 18{\rm{ m/s}}\\\end{array}$

Ans: Part a

The particle’s position at $t = 0.3{\rm{ s}}$ is $x = 54{\rm{ m}}$ .

Part b

The particle’s velocity at $t = 0.3{\rm{ s}}$ is $v = 18{\rm{ m/s}}$ .

Part c

The particle’s acceleration at $t = 0.3{\rm{ s}}$ is $a = - 12{\rm{ m/}}{{\rm{s}}^2}$ .

Part d

The particle’s maximum position along positive x axis is $x = 64{\rm{ m}}$ .

Part e

The particle’s attains maximum position along positive x axis at $t = 4.0{\rm{ s}}$ .

Part h

The particle’s acceleration when $v = 0{\rm{ m/s}}$ is $a = - 24{\rm{ m/}}{{\rm{s}}^2}$ .

Part i

The particle’s average velocity in the given time interval is $18{\rm{ m/s}}$ .