On a two lane road a car A is travelling with a speed of $v = 10 m s^{- 1}$ . Two cars B and C approach car A in opposite directions with a speed u = $15 m s^{- 1}$ . At a certain instant when the B and C are equidistant from A each being l =1000 m, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident with c:

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Solution

The relative velocity of car B with respect to car A,
$V_{B A} = V_{B} - V_{A}$
= 15 - 10 = 5 m/s
Relative velocity of car C with respect to car A,
$V_{C A}$ $= V_{C} - (- V_{A})$
$= 15 + 10 = 25 m / s$
At a certain instance, both care at the same distance from car A i.e.,
s = 1 km = 1000 m
Time is taken (t) by car C to cover 1000 m
$= \frac{1000}{25}$ = 40 s
Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s.
From second equation of motion, minimum acceleration ( $α$ ) produced by car B can be obtained as:
s $= u t + (\frac{1}{2}) a r^{2}$
$1000 = 5 \times 40 + (\frac{1}{2}) \times a \times (40)^{2}$
a $= \frac{1600}{1600} = 1 m s^{2}$

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