Question

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

Answer 1

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

Answer 2

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Velocity of car A,

vA = 36 km/h = 10 m/s

Velocity of car B,

vB = 54 km/h = 15 m/s

Velocity of car C,

vC = 54 km/h = 15 m/s

Relative velocity of car B with respect to car A, vBA = vB – vA = 15 – 10 = 5 m/s

Relative velocity of car C with respect to car A, vCA = vC – (– vA) = 15 + 10 = 25 m/s

At a certain instance, both cars B and C are at the same distance from car A i.e., s = 1 km = 1000 m

Time taken (t) by car C to cover 1000 m = 1000/ 25 = 40

Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s. From second equation of motion, minimum acceleration (a) produced by car B can be obtained as:

S= ut +1/2 at^2

1000= 5 × 40 = 1/2 ×a × (40)^2

a = 1600/1600 = 1 m/s^2

I hope, this will help you

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