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Motion in a Straight Line

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Question : 8 of 27
Marks: +1, -0
On a two-lane road, car A is travelling with a speed of 36 km h−1^{-1}. Two cars B and C approach car A in opposite directions with a speed of 54 km h−1^{-1} each. At a 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?
Solution:  
Velocity of car A=36 km h−1=10 ms−1A=36\ \text{km}\ \text{h}^{-1}=10\ \text{ms}^{-1};
Velocity of car BB or C=54 km h−1=15 ms−1C=54\ \text{km}\ \text{h}^{-1}=15\ \text{ms}^{-1};
Relative velocity of BB w.r.t. A=15−10=5 ms−1A=15-10=5\ \text{ms}^{-1};
Relative velocity of CC w.r.t. A=15+10=25 ms−1A=15+10=25\ \text{ms}^{-1};
As,AB=AC=1 km=1000 m\text{As}, AB=AC=1\ \text{km}=1000\ \text{m}
Time available to BB or CC for crossing A=100025=40 sA=\frac{1000}{25}=40\ \text{s}
If car B accelerates with acceleration a, to cross A before car C does, then
u=5 m s−1,u=5\ \text{m}\ \text{s}^{-1},
t=40 s,t=40\ \text{s},
S=1000 m,S=1000\ \text{m},
a=?a=?
Using, S=ut+12at2,S=ut+\frac{1}{2}at^{2},
we have, 1000=5×40+12a×4021000=5\times40+\frac{1}{2}a\times40^{2}
or   1000−200=800a\;1000-200=800a
or a=1 ms−2a=1\ \text{ms}^{-2}
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