Question

two cars A and B moving in same direction on a straight road with a velocity of 72 km/hour and 60 km per hour respectively when the separation between the two Cars was 2 kilometre car B started accelerating to avoid collision what is the minimum acceleration of car B so that they do not collide​

Answer 1

Answer:

acceleration of car B will be 0.00278 m/s^2 in order to avoid collision

Explanation:

As we know that both cars are moving with different velocities

So here relative velocity of car A with respect to car B is given as

$v_{ab} = v_a - v_b$

$v_{ab} = 72 - 60$

$v_{ab} = 12 km/h$

now we have

$v_{ab} = 12 \times \frac{5}{18} = 3.33 m/s$

now we know that the distance between two cars is

d = 2 km = 2000 m

so here we can say that two cars will remain at rest with respect to each other while the distance between them is zero

so here we have

$v_f^2 - v_i^2 = 2 a d$

$0 - 3.33^2 = 2(a)(2000)$

$a = 2.78 \times 10^{-3} m/s^2$

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Topic : Kinematics

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