Question

3.8. On a two-lane road, car A is travelling with a speed of 36 km h-1. Two cars B and C approach car A in opposite directions with a speed of 54 km h-1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident ?

Answer 1

Car A is in the middle.  Car B and C are on either side of car A.  Let car B travel in the same direction as A.

Va = speed of car A = 36 kmph 
Vb = - 54 kmph = Vc
Relative velocity of c wrt A :  54+36 = 90 kmph
distance between them = 1 km
time to cross = 1/90 hrs

Initial Relative velocity of car B wrt A : 54 - 36 = 18 kmph
max. Time to cross = 1/90  hrs
Distance : 1 km

         s = u t + 1/2 a t²
         1 = 18 / 90 + 1/2 * a /90²
         a = 2*90 * 72 *1000/(3600*3600)  m/s^2 
         a =  1 m/s^2   or  12,960 km/hour^2

Answer 2

Speed of car A=36 km/h

=36*5/18=10 m/s

let Vb and Vc be speed of car B and C

Vb=Vc=54 km/h

=54*5/18

15 m/s

relative speed of car B with respect to car A,Vba is

Vba=Vb-Va=15-10=5m/s

relative speed of car C with respect to car A,Vca is

Vca=Vc+Va=15+10=25m/s

AB=AC=1 km

=1000m

Let t be the time taken by c

s=ut

t=s/u=AC/vca=1000/25=40s

Let a be the accelaration of car B for time t =40 sec

wkt

s=ut+1/2at^2

1000=5*40+1/2a*40*40

1000=200+800a

800=800a

a=1 m/s^2

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