Chapter 11.1, Problem 11.8P

The motion of a particle is defined by the relation $x=t^2-{\left(t-2\right)}^3$ , where x and t are expressed in feet and seconds, respectively. Determine (a) the two positions at which the velocity is zero, (b) the total distance traveled by the particle from $t=0$ to $t=4$s.

To determine

(a)

The positions at which velocity is zero.

Answer to Problem 11.8P

Time $x_1=8.88ftand{x}_2=2.03ft$

Explanation of Solution

Given information:

The motion of the particle is given by the equation:

$x=t^2-{\left(t-2\right)}^3$

$\begin{array}{l}x=t^2-\left(t^3-{(2)}^3-3t\times 2\left(t-2\right)\right)\\ x=t^2-\left(t^3-8-6t^2+12t\right)\\ x=t^2-t^3+8+6t^2-12t\end{array}$

$x=-t^3+7t^2-12t+8$ __________ (1)

Where (x) in feet and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t),

Since, $v=\frac{dx}{dt}$

$\begin{array}{l}v=\frac{d}{dt}\left(-t^3+7t^2-12t+8\right)\\ v=-3t^2+14t-12\end{array}$ __________ (2)

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t)

$\begin{array}{l}a=\frac{dv}{dt}\\ a=\frac{d}{dt}\left(-3t^2+14t-12\right)\end{array}$

$a=-6t+14$ _________ (3)

When velocity is zero, v=0

From equation (2),

$\begin{array}{l}v=-3t^2+14t-12or,\\ 0=-3t^2+14t-12or,\\ -3t^2+14t-12=0\\ t_1=3.53s\\ t_2=1.131s\\ \end{array}$

Now, the position when t₁=3.53s.

$\begin{array}{l}{x}_1=-t^3+7t^2-12t+8\\ x_1=-{(}^{3.53}+7{(}^{3.53}-12(3.53)+8\\ x_1=-43.98+87.22-42.36+8\\ x_1=95.22-86.34\\ x_1=8.88ft\end{array}$

Now, the position when t₁=1.131s:

$\begin{array}{l}{x}_2=-t^3+7t^2-12t+8\\ x_2=-{(}^{1.31}+7{(}^{1.31}-12(1.31)+8\\ x_2=-2.25+12-15.72+8\\ x_2=20-17.97\\ x_2=2.03ft\end{array}$

Conclusion:

The two positions when velocity is zero x₁=8.88 ft and x₂=2.03ft.

To determine

(b)

The total distance traveled by the particle when, t = 0s to t=4s.

Answer to Problem 11.8P

Total distance travel by the particle is 8 ft.

Explanation of Solution

Given information:

The motion of the particle is given by the equation:

$x=t^2-{\left(t-2\right)}^3$

$\begin{array}{l}x=t^2-\left(t^3-{(2)}^3-3t\times 2\left(t-2\right)\right)\\ x=t^2-\left(t^3-8-6t^2+12t\right)\\ x=t^2-t^3+8+6t^2-12t\end{array}$

$x=-t^3+7t^2-12t+8$ __________ (1)

Where (x) in feet and (t) is in seconds.

We can obtain the velocity (v) at any time (t) by differentiating (x) with respect to (t),

Since, $v=\frac{dx}{dt}$

$\begin{array}{l}v=\frac{d}{dt}\left(-t^3+7t^2-12t+8\right)\\ v=-3t^2+14t-12\end{array}$ __________ (2)

The acceleration (a) can be obtained by differentiating again the above equation with respect to (t):

$\begin{array}{l}a=\frac{dv}{dt}\\ a=\frac{d}{dt}\left(-3t^2+14t-12\right)\end{array}$

$a=-6t+14$ _________ (3)

From equation (1), when t = 0s.

$\begin{array}{l}{x}_1=-t^3+7t^2-12t+8\\ x_1=-{(}^0+7{(}^0-12(0)+8\\ x_1=8ft\end{array}$

Again, from equation (1), when t=4s

$\begin{array}{l}{x}_2=-t^3+7t^2-12t+8\\ x_2=-{(}^4+7{(}^4-12(4)+8\\ x_2=-64+112-48+8\\ x_2=120-112\\ x_2=8ft\end{array}$

Conclusion:

The total distance travel by the particle is 8 ft.