On a two lane road, a car $A$ is travelling with a speed of $10 m / s$ . Two cars $B$ and $C$ approach car $A$ in opposite directions with speed of $15 m / s$ each. At an instant, when the distance $A B$ is equal to $A C$ , both being $1 k m, B$ decides to overtake $A$ before $C$ does. What minimum acceleration of car $B$ is required to avoid an accident?

Open in App

Solution

Speed of car A=36 km/h
$= 36 \times 5 / 18 = 10 m / s$
let $V_{b} a n d V_{c}$ be speed of car B and C respectively
$V_{b} = V_{c} = 54 k m / h$
$= 54 \times 5 / 18 = 15 m / s$
relative speed of car B with respect to car $A, V_{b a}$ is
$V_{b a} = V_{b} - V_{a} = 15 - 10 = 5 m / s$
relative speed of car C with respect to car $A, V_{c a}$ is
$V_{c a} = V_{c} + V_{a} = 15 + 10 = 25 m / s$
$A B = A C = 1 k m = 1000 m$
Let t be the time taken by car AC
$s = u t$
$t = s / u = A C / v_{c a} = 1000 / 25 = 40 s$
Let a be the acceleration of car B for time $t = 40 s e c$
$s = u t + 1 / 2 a t^{2}$
$1000 = 5 \times 40 + 1 / 2 a \times 40 \times 40$
$1000 = 200 + 800 a$
$800 = 800 a$
$a = 1 m / s^{2}$

Suggest Corrections

Similar questions

A

36 k m h^{- 1}

B

C

A

54 k m h^{- 1}

A B

A C

1 k m, B

A

C

B

h^{- 1}

h^{- 1}

Join BYJU'S Learning Program