Chapter 11, Problem 11.182RP

Students are testing their new drone to see if it can safely deliver packages to different departments on campus. Position data can be approximated using the expressions $x\left(t\right)=-0.0000225t^4+0.003t^3+0.01t^2$ and $y\left(t\right)=300\left[1-\cos \left(\frac{\pi }{40}t\right)\right]$ , where x and y are expressed in meters and t is expressed in seconds. Knowing that the take-off and landing altitudes are the same, plot the path of the drone and determine (a) the duration of the flight, (b) its maximum speed in the x direction, (c) its maximum altitude and the horizontal distance traveled during the flight.

To determine

(a)

The duration of flight.

Answer to Problem 11.182RP

We got the time of flight is $t=80\text{ sec}$ .

Explanation of Solution

Given information:

Time $t$

$x(t)=-0.0000225t^4+0.003t^3+0.01t^2$

$y(t)=300\left[1-\cos \left(\frac{\pi }{40}t\right)\right]$

Concept used:

We shall draw graph (plot of path)

Calculation:

Following table is made-

t	0.00	1.00	10.00	20.00	30.00	39.00	40.00	50.00	60.00	70.00
x(t)	0.00	0.01	3.78	24.40	71.78	141.11	150.40	259.38	392.40	537.78
y(t)	0.00	0.93	87.89	300.06	512.20	599.08	600.00	512.02	299.82	87.72

t	79.00	80.00	81.00	82.00
x(t)	665.15	678.40	691.38	704.07
y(t)	0.91	0.00	0.94	3.73

Plot,

From above table and plot we get the take off and landing altitudes are the same at

$t=0\text{ and }t=80\text{ sec}$

Hence the time of flight is

$t=80\text{ sec}$

Conclusion:

We got the time of flight is

$t=80\text{ sec}$

To determine

(b)

The maximum speed in x direction.

Answer to Problem 11.182RP

We got the maximum horizontal speed $v_{\max }=14.678\text{ m/s}$

Explanation of Solution

Given information:

Time $t$

$x(t)=-0.0000225t^4+0.003t^3+0.01t^2$

$y(t)=300\left[1-\cos \left(\frac{\pi }{40}t\right)\right]$

Concept used:

Speed $v=\frac{d}{dt}\left[x(t)\right]$

For maximum speed $\frac{dv}{dt}=0$

Calculation:

Speed

$\begin{array}{l}v=\frac{d}{dt}\left[x(t)\right]\\ v=\frac{d}{dt}\left[-0.0000225t^4+0.003t^3+0.01t^2\right]\\ v=-0.00009t^3+0.009t^2+0.02t\end{array}$

For maximum speed

$\begin{array}{l}\frac{dv}{dt}=0\\ \frac{d}{dt}(-0.00009t^3+0.009t^2+0.02t)=0\\ -0.00027t^2+0.018t+0.02=0\\ \text{on solving,}\\ \text{t=67}\text{.75985 sec}\end{array}$

On putting value of $\text{t=67}\text{.75985 }$ in speed we get,

$v=14.678\text{ m/s}$

Conclusion:

We got the maximum horizontal speed $v_{\max }=14.678\text{ m/s}$

To determine

(c)

The maximum altitude

The horizontal distance of the flight.

Answer to Problem 11.182RP

We get the maximum altitude, $H=600\text{ m}$

And horizontal distance traveled, $d=678.3\text{ m}$

Explanation of Solution

Given information:

Time $t$

$x(t)=-0.0000225t^4+0.003t^3+0.01t^2$

$y(t)=300\left[1-\cos \left(\frac{\pi }{40}t\right)\right]$

Concept used:

We shall draw graph (plot of path)

Calculation:

Following table is made-

t	0.00	1.00	10.00	20.00	30.00	39.00	40.00	50.00	60.00
x(t)	0.00	0.01	3.78	24.40	71.78	141.11	150.40	259.38	392.40
y(t)	0.00	0.93	87.89	300.06	512.20	599.08	600.00	512.02	299.82

t	70.00	79.00	80.00	81.00	82.00
x(t)	537.78	665.15	678.40	691.38	704.07
y(t)	87.72	0.91	0.00	0.94	3.73

Plot,

From above table and plot we get the maximum altitude,

$H=600\text{ m}$

And horizontal distance travelled,

$d=678.3\text{ m}$

Conclusion:

We get the maximum altitude, $H=600\text{ m}$

And horizontal distance travelled, $d=678.3\text{ m}$