Chapter 11.1, Problem 11.4P

A loaded railroad car is rolling at a constant velocity when it couples with a spring and dashpot bumper system. After the coupling, the motion of the car is defined by the relation x = 60e^−4.8t sin 16t, where x and t are expressed in millimeters and seconds, respectively. Determine the position, the velocity, and the acceleration of the railroad car when (a) t = 0, (b) t = 0.3 s.

Fig. P11.4

To determine

The position (x), velocity (v), and acceleration (a) of the car when t is 0 seconds.

Answer to Problem 11.4P

The position (x), velocity (v) and acceleration (a) are $\underset{\_}{0\text{mm}}$, $\underset{\_}{960\text{mm/s}}$, and $\underset{\_}{9,216\text{mm}/{\text{s}}^2}$ respectively.

Explanation of Solution

Given information:

The function of time is $x=60e^{-4.8t}\sin 16t$.

Calculation:

Write the relation for the motion of car:

$x=60e^{-4.8t}\sin 16t$ (1).

Here, x is position of car and t is time.

Calculate the position (x) of the car when t is $\text{0}$:

Substitute $\text{0}\text{sec}$ for t in Equation (1).

$\begin{array}{c}x=60e^{-4.8\left(0\right)}\sin 16\left(0\right)\\ =0\text{mm}\end{array}$

Calculate the velocity (v) of the car when t is 0 sec.

Differentiate Equation (1) with respect to time.

$\frac{dx}{dt}=60\left[\left(-4.8\right)e^{-4.8t}\sin 16t+\left(16\right)e^{-4.8t}\cos 16t\right]$

First derivative of position is equal to the velocity of the car. Rewrite the above equation as given below:

$v=60\left[\left(-4.8\right)e^{-4.8t}\sin 16t+\left(16\right)e^{-4.8t}\cos 16t\right]$ (2).

Here, v is velocity of the car.

Calculate the velocity (v):

Substitute 0 sec for t in Equation (2).

$\begin{array}{c}v=60\left[\left(-4.8\right)e^{-4.8\left(0\right)}\sin 16\left(0\right)+\left(16\right)e^{-4.8\left(0\right)}\cos 16\left(0\right)\right]\\ =60\left[0+16\right]\\ =960\text{mm}/\text{s}\end{array}$

Calculate the acceleration (a) of the car when t is 0 sec.

Differentiate Equation (2) with respect to time.

$\begin{array}{c}\frac{dv}{dt}=60\left[\begin{array}{l}\left(-4.8\right)\left(-4.8\right)e^{-4.8t}\sin 16t+\left(-4.8\right)\left(16\right)e^{-4.8t}\cos 16t\\ +\left(-4.8\right)\left(16\right)e^{-4.8t}\cos 16t-\left(16\right)\left(16\right)e^{-4.8t}\sin 16t\end{array}\right]\\ =60\left[\begin{array}{l}23.04e^{-4.8t}\sin 16t-76.8e^{-4.8t}\cos 16t\\ -76.8e^{-4.8t}\cos 16t-256e^{-4.8t}\sin 16t\end{array}\right]\end{array}$ (3).

First derivative of velocity is equal to the acceleration of the car.

Rewrite Equation (3),

$a=60\left[\begin{array}{l}23.04e^{-4.8t}\sin 16t-76.8e^{-4.8t}\cos 16t\\ -76.8e^{-4.8t}\cos 16t-256e^{-4.8t}\sin 16t\end{array}\right]$ (4).

Calculate the acceleration (a) of the car:

Substitute 0 sec for t in Equation (4).

$\begin{array}{c}a=60\left[\begin{array}{l}23.04e^{-4.8\left(0\right)}\sin 16\left(0\right)-76.8e^{-4.8\left(0\right)}\cos 16\left(0\right)\\ -76.8e^{-4.8\left(0\right)}\cos 16\left(0\right)-256e^{-4.8\left(0\right)}\sin 16\left(0\right)\end{array}\right]\\ =60\left[0-76.8-76.8-0\right]\\ =-9,216\text{mm}/{\text{s}}^2\end{array}$

Therefore, the position (x), velocity (v) and acceleration (a) are $\underset{\_}{0\text{mm}}$, $\underset{\_}{960\text{mm/s}}$ and $\underset{\_}{9,216\text{mm}/{\text{s}}^2}$ respectively.

To determine

The position (x), velocity (v), and acceleration (a) of the car when t is 0.3 seconds.

Answer to Problem 11.4P

Therefore, the position (x), velocity (v) and acceleration (a) are $\underset{\_}{14.16\text{mm}}$, $\underset{\_}{87.9\text{mm}/\text{s}}$ and $\underset{\_}{3.11\text{ m}/{\text{s}}^2}$ respectively.

Explanation of Solution

Given information:

The function of time is $x=60e^{-4.8t}\sin 16t$.

Calculation:

Write the relation for the motion of car as given below:

$x=60e^{-4.8t}\sin 16t$

Here, position of car is x in mm and time is t in seconds.

Calculate the position (x) of the car when t is $\text{0}\text{.3}\text{sec}$:

Substitute 0.3s for t in Equation (1).

$\begin{array}{c}x=60e^{-4.8\left(0.3\right)}\sin 16\left(0.3\right)\\ =60\left(0.2367\right)\left(-0.9962\right)\\ =14.16\text{mm}\end{array}$

Calculate the velocity (v) of the car when t is $\text{0}\text{.3}\text{sec}$:

Substitute 0.3s for t in Equation (3).

$\begin{array}{c}v=60\left[\left(-4.8\right)e^{-4.8\left(0.3\right)}\sin 16\left(0.3\right)+\left(16\right)e^{-4.8\left(0.3\right)}\cos 16\left(0.3\right)\right]\\ =60\left[1.1328+0.3317\right]\\ =60\left[1.4645\right]\\ \approx 87.9\text{mm}/\text{s}\end{array}$

Calculate the acceleration (a) of the car when t is $\text{0}\text{.3}\text{sec}$.

Substitute 0.3s for t in Equation (3).

$\begin{array}{c}a=60\left[\begin{array}{l}23.04e^{-4.8\left(0.3\right)}\sin 16\left(0.3\right)-76.8e^{-4.8\left(0.3\right)}\cos 16\left(0.3\right)\\ -76.8e^{-4.8\left(0.3\right)}\cos 16\left(0.3\right)-256e^{-4.8\left(0.3\right)}\sin 16\left(0.3\right)\end{array}\right]\\ =60\left[-5.4372-1.5919-1.5919+60.4138\right]\\ =3,107.568\left(\frac{\text{m}/{\text{s}}^2}{1,000\text{mm}/{\text{s}}^2}\right)\\ \approx 3.11\text{ m}/{\text{s}}^2\end{array}$

Therefore, the position (x), velocity (v) and acceleration (a) are $\underset{\_}{14.16\text{mm}}$, $\underset{\_}{87.9\text{mm}/\text{s}}$ and $\underset{\_}{3.11\text{ m}/{\text{s}}^2}$ respectively.