Chapter 11, Problem 11.182RP

(a)

To determine

The time (t) when the velocity is zero.

Answer to Problem 11.182RP

The time (t) when the velocity is zero are $\underset{\_}{1\sec }$ and $\underset{\_}{4\sec }$.

Explanation of Solution

Given information:

The motion of the particle is defined by the relation $x=2t^3-15t^2+24t+4$.

Calculation:

Write the relation for the model car:

$x=2t^3-15t^2+24t+4$ (1).

Differentiate Equation (1) with respect to time.

$v\left(t\right)=\frac{dx}{dt}=6t^2-30t+24$ (2).

Here, velocity of the particle is v.

Differentiate Equation (2) with respect to time.

$a\left(t\right)=\frac{dv}{dt}=12t-30$ (3).

Here, acceleration of the particle is a.

Substitute 0 for v in Equation (2).

$\begin{array}{l}0\left(t\right)=6t^2-30t+24\\ 0=6\left(t^2-5t+4\right)\\ 0=\left(t-4\right)\left(t-1\right)\end{array}$

Therefore, the time (t) when the velocity is zero are $\underset{\_}{1\sec }$ and $\underset{\_}{4\sec }$.

(b)

To determine

The position (x) and total distance (d) travelled when the acceleration is zero.

Answer to Problem 11.182RP

The position (x) and total distance (d) are $\underset{\_}{1.50\text{m}}$ and $\underset{\_}{24.5\text{m}}$ respectively.

Explanation of Solution

Given information:

The motion of the particle is defined by the relation $x=2t^3-15t^2+24t+4$.

Calculation:

Calculate the time $\left(t\right)$ when a is $\text{0}\text{sec}$ using equation (3).

Substitute 0 for a in Equation (3).

$\begin{array}{l}0\left(t\right)=12t-30\\ -12t=-30\\ t=2.5\sec \end{array}$

For time $0\le t\le 1\text{s}$ the acceleration is $a>0$.

For time $1\text{s}\le t\le 2.5\text{s}$ the acceleration is $a\le 0$.

Calculate the position $\left(x_0\right)$ of the particle when t is 0 sec:

Substitute 0 sec for t in Equation (1).

$\begin{array}{c}{x}_0=2{\left(0\right)}^3-15{\left(0\right)}^2+24\left(0\right)+4\\ =\text{4}\text{m}\end{array}$

Calculate the position $\left(x_1\right)$ of the particle when t is 1 sec:

Substitute 1 sec for t in Equation (1).

$\begin{array}{c}{x}_1=2{\left(1\right)}^3-15{\left(1\right)}^2+24\left(1\right)+4\\ =2-15+24+4\\ =15\text{m}\end{array}$

Calculate the position $\left(x_2\right)$ of the particle when t is 2.5 sec:

Substitute 2.5 sec for t in Equation (1).

$\begin{array}{c}{x}_2=2{\left(2.5\right)}^3-15{\left(2.5\right)}^2+24\left(2.5\right)+4\\ =31.25-93.75+60+4\\ =1.5\text{m}\end{array}$

Calculate the distance travelled $\left(d_1\right)$ when time (t) is from 0 sec to 1sec using the relation:

$d_1=\left|x_1-x_0\right|$

Substitute 4 m for $x_0$ and 15m for $x_1$.

$\begin{array}{c}{d}_1=\left|15-4\right|\\ =11\text{m}\end{array}$

Calculate the distance travelled $\left(d_2\right)$ when time (t) is from 1 sec to 2.5 sec using the relation:

$d_2=\left|x_2-x_1\right|$

Substitute 15m for $x_1$ and 1.5 m for $x_2$.

$\begin{array}{c}{d}_2=\left|1.5-15\right|\\ =13.5\text{m}\end{array}$

Calculate the total distance (d) using the relation:

$d=d_1+d_2$

Substitute 11 m for $d_1$ and 13.5 for $d_2$.

$\begin{array}{c}d=11+13.5\\ =24.5\text{m}\end{array}$

Therefore, the position (x) and total distance (d) are $\underset{\_}{1.50\text{m}}$ and $\underset{\_}{24.5\text{m}}$ respectively.