Average spacing between cars of a row = [tex]120\textrm{ }m[/tex]
Explanation:
Let us assume the average spacing between cars to be [tex]S[/tex].
Frame of Reference : Frame attached to Cars moving at [tex]60 \frac{km}{h}[/tex].
In this frame, the cars in the row are at rest, while the car travelling in opposite direction appears to move at a relative speed of [tex]60 \frac{km}{h} +80 \frac{km}{h}=140\frac{km}{h}[/tex].
He is covering a distance of [tex]1\textrm{ }km[/tex] in ground frame. His speed in ground frame is [tex]80 \frac{km}{h}[/tex]. So, time taken to traverse 1 km is
[tex]Time taken = \frac{\textrm{Distance covered in that frame}}{\textrm{Speed in that frame }}[/tex]
[tex]T = \frac{1 km}{80 \frac{km}{h} }[/tex]
[tex]T = \frac{1}{80} h[/tex]
In this T time, he moves in the relative frame at [tex]140[/tex] [tex]\frac{km}{h}[/tex]. The distance covered in relative frame will be
[tex]\textrm{Distance covered in a frame} = \textrm{Speed in that frame}[/tex]×[tex]\textrm{time}[/tex]
[tex]Distance = \textrm{140 } \frac{km}{h}[/tex] × [tex]\frac{1}{80} h[/tex]
[tex]Distance = \frac{7}{4} km[/tex]
It passes 14 cars while traversing this distance.
∴ [tex]\frac{7}{4} km =[/tex] 14 × [tex](\textrm{Length of a car + Distance between two cars})[/tex]
[tex]\frac{7}{4} km = 14[/tex] × [tex](5 m+S)[/tex]
[tex]\frac{1}{8}\textrm{ }km\textrm{ } =\textrm{ } 5\textrm{ }m\textrm{ }+\textrm{ } S[/tex]
[tex]S\textrm{ }=\textrm{ }125\textrm{ }m\textrm{ }-\textrm{ }5\textrm{ }m\textrm{ }=\textrm{ }120\textrm{ }m[/tex].
∴Average Spacing betwwen cars in a row = [tex]120\textrm{ }m[/tex]
