Question

You are kidnapped by political-science majors (who are upset because you told them political science is not real science). Although blindfolded, you can tell the speed of their car (by the whine of the engine), the time of travel (by mentally counting off seconds), and the direction of travel (by turns along the rectangular street system). From these clues, you know that you are taken along the following course: $\mathbf{50}\mathbf{Km}\mathbf{/}\mathbf{h}$for$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{min}$, turn $\mathbf{90}\mathbf{°}$to the right,$\mathbf{20}\mathbf{Km}\mathbf{/}\mathbf{h}$ for$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{min}$, turn $\mathbf{90}\mathbf{°}$to the right, $\mathbf{20}\mathbf{Km}\mathbf{/}\mathbf{h}$for$\mathbf{60}\mathbf{s}$, turn $\mathbf{90}\mathbf{°}$to the left,$\mathbf{50}\mathbf{Km}\mathbf{/}\mathbf{h}$for$\mathbf{60}\mathbf{s}$, turn $\mathbf{90}\mathbf{°}$to the right, $\mathbf{20}\mathbf{Km}\mathbf{/}\mathbf{h}$for$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{min}$, turn$\mathbf{90}\mathbf{°}$ to the left $\mathbf{50}\mathbf{Km}\mathbf{/}\mathbf{h}$for$\mathbf{30}\mathbf{s}$. At that point, (a) how far are you from your starting point, and (b) in what direction relative to your initial direction of travel are you?

Short answer

a) The distance of a kidnapped point from the starting point is $2668m$.

b) The direction of travel from the initial point$-76°$i.e. measured clockwise with respect to the initial direction of motion.

Step-by-step solution

The given data

Speed and the direction of movement along the following course are given as $50km/h$for$2min$, turn $90°$to the right$20km/h$, for$4min$, turn60s, turn $90°$to the left, $50km/h$for $60s$, turn $90°$to the right, $20km/h$for$2min$, turn $90°$to the left, $50km/h$for$30s$.

Understanding the concept of the relative motion

The comparison of the relative accelerations and velocities of two rigid entities is referred to as relative motion. Using the formula for the magnitude of the vector, we can find the magnitude of the displacement and direction of the ending point from the starting point.

Formulae:

Magnitude of the resultant vector$\overrightarrow{V}$is given by,$V=\sqrt{V_x^2+V_y^2}$ …(i)

The direction of vector is given by, $\tan \theta =\frac{V_y}{V_x}$ …(ii)

(a) Calculation of the distance of the kidnapped point from the starting point

Distance covered in 2 min at a speed of 50 km/h

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}1}=50\frac{\mathrm{km}}{\mathrm{hr}}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1\mathrm{hr}}{60\min }\times \frac{1\min }{60\mathrm{s}} \\ =13.89\mathrm{m}/\mathrm{s} \\ {\mathrm{t}}_1=2\min \times \frac{60}{1\min } \\ =120\mathrm{s} \end{gathered}$$

So, the distance $\left(d_1\right)$is

$$\begin{gathered} d_1=v_{x1}\times t_1 \\ =13.89m/s\times 120s \\ =1666.67 \\ \approx 1667m \end{gathered}$$

The distance traveled $\left(d_2\right)$in 4 min at a speed of 20 km/h

$$\begin{gathered} {\mathrm{v}}_{y1}=20\frac{\mathrm{km}}{\mathrm{hr}}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1\mathrm{hr}}{60\min }\times \frac{1\min }{60\mathrm{s}} \\ =5.56\mathrm{m}/\mathrm{s} \end{gathered}$$

$$\begin{gathered} {\mathrm{t}}_2=4\min \times \frac{60}{1\min } \\ =240\mathrm{s} \end{gathered}$$

So, the distance $\left(d_2\right)$is

$$\begin{gathered} d_2=v_{y1}\times t_2 \\ =5.56m/s\times 240s \\ =1333.33 \\ \approx 1333m \end{gathered}$$

The distance traveled $\left(d_3\right)$in 60 sec at a speed of 20 km/h

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}2}=20\frac{\mathrm{km}}{\mathrm{hr}}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1\mathrm{hr}}{60\min }\times \frac{1\min }{60\mathrm{s}} \\ =5.56\mathrm{m}/\mathrm{s} \\ {\mathrm{t}}_3=60s \end{gathered}$$

So, the distance $\left(d_3\right)$is

$$\begin{gathered} d_3=v_{x2}\times t_3 \\ =5.56m/s\times 60s \\ =333.33m \\ \approx 333m \end{gathered}$$

Distance covered $\left(d_4\right)$in 60 s at a speed of 50 km/h

$$\begin{gathered} {\mathrm{v}}_{y2}=50\frac{\mathrm{km}}{\mathrm{hr}}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1\mathrm{hr}}{60\min }\times \frac{1\min }{60\mathrm{s}} \\ =13.89\mathrm{m}/\mathrm{s} \\ {\mathrm{t}}_4=60s \end{gathered}$$

So, the distance$\left(d_4\right)$is

$$\begin{gathered} d_4=v_{y2}\times t_4 \\ =13.89m/s\times 60s \\ =833.33m \\ \approx 833m \end{gathered}$$

Distance covered $\left(d_5\right)$in 2 min at a speed of 20 km/h

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}3}=20\frac{\mathrm{km}}{\mathrm{hr}}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1\mathrm{hr}}{60\min }\times \frac{1\min }{60\mathrm{s}} \\ =5.56\mathrm{m}/\mathrm{s} \\ {\mathrm{t}}_5=2\min \times \frac{60\mathrm{s}}{1\min } \\ =120\mathrm{s} \end{gathered}$$

So, the distance $\left(d_5\right)$is

$$\begin{gathered} {\mathrm{d}}_5={\mathrm{v}}_{\mathrm{x}3}\times {\mathrm{t}}_5 \\ =5.56\mathrm{m}/\mathrm{s}\times 120\mathrm{s} \\ =667.2\mathrm{m} \\ \approx 667\mathrm{m} \end{gathered}$$

Distance covered$\left(d_6\right)$in 30 s at a speed of 50 km/h

$$\begin{gathered} {\mathrm{v}}_{y3}=50\frac{\mathrm{km}}{\mathrm{hr}}\times \frac{1000\mathrm{m}}{1\mathrm{km}}\times \frac{1\mathrm{hr}}{60\min }\times \frac{1\min }{60\mathrm{s}} \\ =13.89\mathrm{m}/\mathrm{s} \\ {\mathrm{t}}_6=30s \end{gathered}$$

So, the distance $\left(d_6\right)$is

$$\begin{gathered} d_6=v_{y3}\times t_6 \\ =13.89m/s\times 30s \\ =416.66m \\ \approx 417m \end{gathered}$$

The distances traveled can be shown in the below diagram. The direction towards right and the upward direction is assumed as positive.

From diagram and calculated values, we can find the net displacement along x-direction as,

$$\begin{gathered} X=d_1-d_3-d_5 \\ =1667m-333m-667m \\ =667m \end{gathered}$$

From diagram and calculated values, we can find the net displacement along y-direction as,

$$\begin{gathered} \mathrm{Y}={\mathrm{d}}_2+{\mathrm{d}}_4+{\mathrm{d}}_6 \\ =\left(-1333\mathrm{m}\right)+\left(-833\mathrm{m}\right)+\left(-417\mathrm{m}\right) \\ =-2583\mathrm{m} \end{gathered}$$

Therefore, the magnitude of total displacement is given using equation (i) and the above components as follows:

$$\begin{gathered} \mathrm{d}=\sqrt{{\left(667\mathrm{m}\right)}^2+{\left(-2583\mathrm{m}\right)}^2} \\ =2667.72\mathrm{m} \\ \approx 2668\mathrm{m} \end{gathered}$$

Hence, the magnitude of the distance vector is$2668m$.

(b) Calculation of the direction of the travel from the initial point

The direction of travel from the initial point is given using equation (ii) as follows:

$$\begin{gathered} \tan \theta =\left(\frac{-2583m}{667m}\right) \\ \theta ={\tan }^{-1}\left(-3.873\right) \\ =-75.52° \\ \approx -76° \end{gathered}$$

Since angle is negative, it means it is measured in clockwise direction with respect to initial direction of motion. Hence, the value of the angle is$76°$.