NEET Physics Question - Kinematics

Two cars A and B are approaching each other head-on with speeds 20 m/s and 10 m/s respectively. When their separation is X then A and B start braking at 4 m/s² and 2m/s² respectively. Minimum value of X to avoid collision is

60 m

75 m

80 m

90 m

Solution:

Using the stopping distance formula:

$s = \frac{u^2}{2a}$

For Car A (

$u_A = 20$

m/s,

$a_A = 4$

m/s²):

$s_A = \frac{20^2}{2 \times 4} = \frac{400}{8} = 50 \text{ m}$

For Car B (

$u_B = 10$

m/s,

$a_B = 2$

m/s²):

$s_B = \frac{10^2}{2 \times 2} = \frac{100}{4} = 25 \text{ m}$

Total minimum separation to avoid collision:

$X = s_A + s_B = 50 + 25 = 75 \text{ m}$

Rankers Physics

Solution:

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