On a highway, two buses A and B are running at the same velocity of magnitude 30 ms-1. The brakes caused a deceleration of 307ms−2 in bus A and that of bus B is 3 ms-2. In an emergency when driver of the front car applies brakes, immediately its rear light turns red and braking begins. In response, driver of the rear bus also applies brakes to avoid a collision with the front bus. Every driver takes 1 s to apply the brakes after he saw a need for it. If bus A ahead of bus B, then the minimum separation between the buses before driver of bus A applies the brake is x. If bus B is running ahead of bus A, then the minimum separation between the buses before the driver of bus B applies brake is x2. The value of x13x2is _______
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answer is 5.
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Detailed Solution
For x1, If collision does not like place. The final velocities of both should be same.∴ vf=30−307t=30−3(t−1)∴ 307t=3t−3 or 307t−3t=−3 Or 9t7=−3⇒t=−219The negative of t indicates that both come to rest before collision. The distance moved by bus B before coming to rest iss1=30×1+3022×3=30+150=180mThe distance moved by bus A before coming to rest is s2=3022×307=30×72=105m∴ x1=s1−s2=180−105=75m For x2,vf=30−307(t−1)=30−3t Or 307t−307=3t or 97t=307∴ t=103sThe positive of t indicates that before collision, velocities of both buses are non-zero but same.So, the distance moved by A before collision iss1=30×1+30×103−1−12×307103−12s1=2653ms2=30×103−12×3×1032=100−503=2503x2=s1−s2=2653−2503=153=5m Or x13x2=753×5=5