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Mon to Sat - 10 AM to 7 PM Asked by preetiagrawal1979 14th May 2021, 6:43 AM
Let  city-x be the origin of cartesian coordinate system. Car-B starts half an hour after Car-A .
Let us assume our time t = 0 , when car-B starts to move.

Since car-A is moving at velocity 40 km/hr towards west , Coordinate of car-A at time t = 0 is given as

At t = 0 , coordinates of car-A = ( -40 × (1/2) km , 0 ) = ( -20 km, 0 )

Coordinate of car-A at time t = [ ( -20 - 40t ) km , 0 ] ..............................(1)

City-y is 30 km from city-x in the direction South-West

At t = 0 , coordinate of car-B = [ - ( 30 / √2 ) km , - ( 30 / √2 ) km ] = ( -21.213 km , -21.213 km )

Car-B is travelling with velocity 20√2 km/hr in north west direction .
Hence velocity components of car-B are given by

vBx = - 20 km/hr  ;  vBy = 20 km/hr

Coordinate of car-B at time t = [ - (21.213 + 20 t ) km , - (21.213 - 20 t ) km ]  .......................... (2)

square of the distance S2 between car-A and car-B at any time t is determined from
the coordinates given in (1) and (2) as follows

S2 = { (1.213 -20t )2 + ( 21.213 - 20t )2 } ................................. (4)

To get the minimum distance between car-A and car-B , we differentiate above expression and
equate to zero to get the time t when the distance between the cars is minimum. after simplification we get , t = 0.56 hour
Minimum distance Smin is determined from eqn.(4) by substituting t = 0.56 hour

we get, Smin = 14.14 km
-------------------------------------------------

Hence minimum distance between cars is not 20 km . Minimum distance is not at 1/(2√2) = 0.354 hour .
Cars will never strike at each other because we get Smin > 0

Answer to question :- (d) none of above

Answered by Expert 14th May 2021, 3:01 PM
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Tags: distance