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BITSAT 2019 Paper

Section: Physics

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Question : 35 of 150

Marks: +1, -0

A man drives a car from station B towards station A at speed 60 km/h. A car leaves station A for station B every 10 min. The distance between A and B is 60 km. The car travels at the speed of 60 km/h. A man drives a car from B towards A at speed of 60 km/h. If he starts at the moment when first car leaves the station B, then how many cars would be meet on the route ?

4
7
9
12

Solution:

(b) Distance between two cars leaving from the station A is,

d =

\frac{1}{6}

x 60 = 10 km

Man meets the first car after time,

t_1 = \frac{60}{60+60} = \frac{1}{2}

h

He will meet the next car after time,

t_2 = \frac{10}{60+60} = \frac{1}{12}

h

In the remaining half an hour, the number of cars he will meet again is, n =

\frac{\frac{1}{2}}{\frac{1}{12}}

= 6

∴ Total number of cars would be meet on route will be 7.

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