on a two Lane road car a is travelling at a speed of 36 km per hour to cars b and c in opposite directions approach a with a speed of 54 km per hour is at an instant when the distance a b 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
Answers
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
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 car AC
w.k.t
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