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two cars A and B are approaching Each Other head on with speed 20 metre per second and 10 metre per second respectively their separation is X then A and B start breaking at 4 metre per second square into metre per second square respectively minimum value of X to avoid collision is |
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Answer» Dear Students, ● Answer - X = 62.5 m ◆ Explanation - Let s1 & s2 be stopping distances for CAR A & car B respectively. Applying Newton's 3rd law of kinematics to car A, v1² = u1² + 2as 0 = 20² + 2×(-4) × s s1 = 400 / 8 s1 = 50 m Applying Newton's 3rd law of kinematics to car B, v2² = u2² + 2as 0 = 10² + 2×(-4) × s s2 = 100 / 8 s2 = 12.5 m Therefore, minimum separation between cars at the time of breaking is - X = s1 + s2 X = 50 + 12.5 X = 62.5 m Hope I was useful.. |
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