Solved Problem on One-dimensional Motion
advertisement   



A car with a length of 3 m travels at a speed of 36 km/h. Calculate the exposure time to shoot the automobile in a 1.5 cm length film, knowing that the image remains reasonably clear it should not move more than the distance d = 0.1 mm on the negative.


Problem data:
  • Car length:    C = 3 m;
  • Speed ​​of the car:    v = 36 km/h;
  • Image length:    c = 1.5 cm;
  • Image displacement:    d = 0.1 mm.
Problem diagram:

During the exposure time tex of the film, the car moves a distance of D, so the image does not miss the sharpness it can not move more than a distance d equal to 0.1 mm (Figure 1).

Figure 1

Solution

First, we will convert the car speed given in kilometers per hour (km/h) to meters per second (m/s), the length of the image given in centimeters (cm) to meters (m), and the displacement of the image given in millimeters (mm) for meters (m) used in the International System of Units (SI).
\[ \begin{gather} v=36\;\frac{\cancel{\text{km}}}{\text{h}}\times \frac{1000\;\text{m}}{1\;\cancel{\text{km}}}\times \frac{1\;\text{h}}{3600\;\text{s}}=10\;\text{m/s}\\[10pt] c=1.5\;\cancel{\text{cm}}\times \frac{1\;\text{m}}{100\;\cancel{\text{cm}}}=\frac{1.5}{10^{2}}\;\text{m}=1.5\times 10^{-2}\;\text{m}\\[10pt] d=0.1\;\cancel{\text{mm}}\times \frac{1\;\text{m}}{1000\;\cancel{\text{mm}}}=\frac{1\times 10^{-1}}{10^{3}}\;\text{m}=1\times 10^{-1}\times 10^{-3}\;\text{m}=1\times 10^{-4}\;\text{m} \end{gather} \]
From Figure 1 we have, the ratio between the size of the car, C, and the size of the image, c, should be the same as, the ratio between the displacement of the car, D, and the displacement of the image, d.
\[ \begin{gather} \frac{C}{c}=\frac{D}{d} \tag{I} \end{gather} \]
Car displacement during exposure time will be
\[ \bbox[#99CCFF,10px] {\bar v=\frac{\Delta S}{\Delta t}} \]
As the car travels with constant speed this is your average speed v=vm, the displacement of the car is ΔS=D, and the interval of time is equal to photo exposure time Δt = tex
\[ \begin{gather} v=\frac{D}{t_{ex}}\\ D=v t_{ex} \tag{II} \end{gather} \]
substituting the expression (II) into expression (I) and the problem data
\[ \begin{gather} \frac{C}{c}=\frac{v t_{ex}}{d}\\ t_{ex}=\frac{C}{c}\frac{d}{v}\\ t_{ex}=\frac{3}{1.5\times 10^{-2}}\times\frac{1\times 10^{-4}}{10}\\ t_{ex}=2\times 10^{2}\times 1\times 10^{-4}\times 10^{-1}\\ t_{ex}=2\times 10^{-3}\;\text{s} \end{gather} \]
\[ \bbox[#FFCCCC,10px] {t_{ex}=0.002\;\text{s}} \]
advertisement   

Creative Commons License
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .