## Question

Car *A* is moving with a speed of 36 km h^{–1} on a two-lane road. Two cars*B* and *C*, each moving with a speed of 54 km h^{–1} in opposite direction on the other lane are approaching car *A*. At a certain instant when the distance *AB* = distance *AC* = 1 km, the diver of car *B* decides to overtake*A* before *C* does. What must be the minimum acceleration of car *B* so as to avoid an accident?

### Solution

1 ms^{–2}

Let us suppose that cars *A* and *B* are moving in the positive *x*-direction. Then car *C* is moving in the negative *x*-direction. Therefore,

. The relative velocity*B* with respect to A is . The relative velocity of C with respect to A is . At time *t* = 0, the distance between *A* and *C* = 1 km = 1000 m. The car *C* will cover a distance *AC* = 1000 m and just reach car *A* at a time t given by

= 40 *s*

Car *B* will overtake car *A* just before car *C* does and avoid an accident, if it acquires a minimum acceleration *a* such that it covers a distance *s* = *AB*= 1000 m in time *t* = 40 *s*, travelling at a relative speed. Putting these values in relation

.

Which gives *a* = 1 ms^{–2}.

#### SIMILAR QUESTIONS

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