Question

The Lost Drone. You and your team are exploring the edge of an Antarctic mountain range and you send a drone ahead to help navigate. After takeoff you lose sight of the drone and, a few seconds later, the controls malfunction and the drone stops sending visual images and navigational information except for speed and directional data. Changing speeds erratically, the drone heads west until it makes a drastic turn at the 5-minute mark to β = 45° east of south. After nearly ten minutes, the speed drops to zero and the drone stops sending data. It has crashed. Using the speed/directional data, the team draws up the graph shown below. (a) How far is the drone from you? (b) In what direction must you go to retrieve it? Express your result as a geographical direction west of south.

Velocity v (m/s) WEST B° East of South TO 100 200 300 Time t(s) 400 500 600

Answer 1

A handwritten solution for this question is given below

4375 Lust AOB tand - 0A Area under velocith us time gor abh = dictance. Area under west er А те 4 дод) - 1x ka -x ba 12:5+ / x7.5x50 425 1G = 187.5 m 7.5 2.5 Ag CD 2 %x5x50 125 m loo 100 Area of box A PEF - * *2-5*50 = 62.5m Rest of the graph has are as a 32x 732 2.5x 50 = 4ooom & Total distance in west a 4375m same method area under & east of south = 2812.5 0= 2812.5 & 8A = 2812-5 sinus's 1988.73 m CA 2 2812.5 Cosus = 1988. 73 m. 2. AB2 4375–1988.73 = 2386.3 m @ distance from drone - SABT+ DAT OB 2 3106.33 m & Otai ( 1 ) 2 39.810 ( south of west

Now direction in West of South = (90°- 39.81) = 50.19

Answer 2

Segment 1: d = vt = (8.0 m/s)(50 s) = 400 m

Segment 2: The kinematic equation that we need is:

${\displaystyle v^2=v_0^2+2ax\Rightarrow x=\frac{v^2-v_0^2}{2a}}$

where we can readily read v and v0 from the graph, but need to calculate a from the graph using

${\displaystyle a=\frac{v-v_0}{t}=\frac{14.0\text{m}/\text{s}-8.0\text{m}/\text{s}}{50.0\text{s}}=0.12\text{m}/{\text{s}}^2}$

so that

${\displaystyle x=\frac{v^2-v_0^2}{2a}=\frac{{\left(14.0\text{m}/\text{s}\right)}^2-{\left(8.0\text{m}/\text{s}\right)}^2}{2\left(0.12\text{m}/{\text{s}}^2\right)}=\begin{array}{}\\ 550\text{\hspace{0.17em}m}\\ \end{array}}$

Segment 3: d = vt = (14.0 m/s)(100 s) = 1400 m

Segment 4:

${\displaystyle a=\frac{v-v_0}{t}=\frac{10.0\text{m}/\text{s}-14.0\text{m}/\text{s}}{50.0\text{s}}=-0.08\text{m}/{\text{s}}^2}$

${\displaystyle x=\frac{v^2-v_0^2}{2a}=\frac{{\left(10.0\text{m}/\text{s}\right)}^2-{\left(14.0\text{m}/\text{s}\right)}^2}{2\left(-0.08\text{m}/{\text{s}}^2\right)}=\begin{array}{}\\ 600\text{\hspace{0.17em}m}\\ \end{array}}$

Segment 5:

${\displaystyle a=\frac{v-v_0}{t}=\frac{12.0\text{m}/\text{s}-10.0\text{m}/\text{s}}{50.0\text{s}}=0.04\text{m}/{\text{s}}^2}$

${\displaystyle x=\frac{v^2-v_0^2}{2a}=\frac{{\left(12.0\text{m}/\text{s}\right)}^2-{\left(10.0\text{m}/\text{s}\right)}^2}{2\left(0.04\text{m}/{\text{s}}^2\right)}=\begin{array}{}\\ 550\text{\hspace{0.17em}m}\\ \end{array}}$

Segment 6:

${\displaystyle a=\frac{v-v_0}{t}=\frac{6.0\text{m}/\text{s}-12.0\text{m}/\text{s}}{50.0\text{s}}=-0.12\text{m}/{\text{s}}^2}$

${\displaystyle x=\frac{v^2-v_0^2}{2a}=\frac{{\left(6.0\text{m}/\text{s}\right)}^2-{\left(12.0\text{m}/\text{s}\right)}^2}{2\left(-0.12\text{m}/{\text{s}}^2\right)}=\begin{array}{}\\ 450\text{\hspace{0.17em}m}\\ \end{array}}$

Segment 7: d = vt = (6.0 m/s)(100 s) = 600 m

Segment 8:

${\displaystyle a=\frac{v-v_0}{t}=\frac{14.0\text{m}/\text{s}-6.0\text{m}/\text{s}}{50.0\text{s}}=0.16\text{m}/{\text{s}}^2}$

${\displaystyle x=\frac{v^2-v_0^2}{2a}=\frac{{\left(14.0\text{m}/\text{s}\right)}^2-{\left(6.0\text{m}/\text{s}\right)}^2}{2\left(0.16\text{m}/{\text{s}}^2\right)}=\begin{array}{}\\ 500\text{\hspace{0.17em}m}\\ \end{array}}$

Segment 9: d = vt = (14.0 m/s)(50.0 s) = 700 m

Segments 1–5 point west and sum to 3500 m. Segments 6–9 point 40° east of south and sum to 2250 m. These can be expressed as the two vectors, ${\displaystyle {\overrightarrow{D}}_1}$ and ${\displaystyle {\overrightarrow{D}}_2,}$ respectively. The final step is to resolve the vectors into their components and add them to get the resultant ${\displaystyle \overrightarrow{R}.}$

D1x = –3500 m

D2x = (2250 m)sin40° = 1446.3 m

D2y = -(2250 m)cos40° = –1723.6 m

The components of the resultant are

Rx = D1x + D2x = (–3500 m) + 1446.3 m = -2053.7 m (reminder: -x direction is west)

Ry = D2y = -1723.6 m (reminder: -y direction is south)

The magnitude of the final displacement vector is  2680

0
${\displaystyle \mathrm{268026R}=\sqrt{R_{\text{x}}^2+R_{\text{y}}^2}=\sqrt{{\left(-2053.7\text{\hspace{0.17em}m}\right)}^2+{\left(-1723.6\text{\hspace{0.17em}m}\right)}^2}=\begin{array}{|c|}\hline 2680\text{\hspace{0.17em}m}\\ \hline\end{array}}$

The angle between the resultant and the south (-y) axis is 50

${\displaystyle \theta ={\tan }^{-1}\left(\frac{R_{\text{x}}}{R_{\text{y}}}\right)={\tan }^{-1}\left(\frac{2053.7\text{\hspace{0.17em}m}}{1723.6\text{\hspace{0.17em}m}}\right)=\begin{array}{|c|}\hline 50.0°\\ \hline\end{array}}$