Question

A car leaves regularly from point $A$ for point $B$ every $10min$. The distance between $A$ and $B$ is $60km$. The car travel at a speed for $60km/h$. Find the number of cars that a man driving from $B$ to $A$ will meet enroute if he starts from $B$ simultaneously with one of the cars leaving $A$. The car from B travels at a speed of $60km/h$.

Answer 1

Time taken for car to travel from $B$ to $A$ is:

$T=\frac{s}{v}=\frac{60}{60}=1hr=60min$

Distance between any two consecutive cars leaving from A is constant and is given by:

$s=10/60\times 60=10km$

Relative speed between any car leaving from $A$ and car leaving from $B$ is $v_r=60-(-60)=120km/hr$

Hence, on crossing one car, next car is met in time,

$t=\frac{s}{v}=\frac{10}{120}hr=5min$

Hence, total number of cars met are:

$n=\frac{T}{t}=12$

This excludes the first car met at $B$. Hence, car starting from $B$ meets a total of $13cars$.