On a two lane road, car A is travelling with a speed of 36km/h . two cars B and C approach car A in opposite directions with a speed of 54km/h each. At 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 ?

Suppose car A, B and C are moving with speeds V_{A}, V_{B} and V_{C} respectively. Let the positive direction of X-axis be taken in the direction of motion of car A and car B. Now V_{A}= 36km/h and V_{B}=54km/h and V_{C}= -54km/h.

Now the relative velocity of car B with car A is

V_{BA}= V_{B}-V_{A} = 54 – 36= 18km/h

Relative velocity of car C with respect to car A is

V_{CA} = V_{C}- V_{A} = -54-36 = -90km/h

Negative sign shows V_{CA} is along negative X- axis.

Now suppose the car C takes time t to cover the distance BA of 1km in slightly less than (1/90)h at a speed V_{BA}= 18km/h

Substituting u= 18km/h , t=1/90 h and S= 1 km in the formula

S= ut +1/2at^{2}

We have 1000= 5x40 + ½ a x (40)^{2}

800a = 1000 – 200

a = 800/800 =1m/s^{2}

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