Ncert Exercises & Solutions

3.8 On a two-lane road, car A is travelling with a speed of 36 km h^–1. Two cars B and C approach car A in opposite directions with a speed of 54 km h^–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 ?

Velocity of car A, v $_{A}$ = 36 km/h = 10 m/s

Velocity of car B, v $_{B}$ = 54 km/h = 15 m/s

Velocity of car C, v $_{C}$ = 54 km/h = 15 m/s

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 cars B and C are at the same distance from car A i.e., s = 1 km = 1000 m

Time 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 the second equation of motion, minimum acceleration (a) produced by car B can be obtained as:

$s = u t + \frac{1}{2} a t^{2}$
$1000 = 5 \times 40 + \frac{1}{2} \times a \times {(40)}^{2}$
$\Rightarrow a = \frac{1600}{1600} = 1 m / s^{2}$

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