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? what is the speed of car B is deducted from A and speed of car C is added with A?
Answers
1 m/s^2
Given:
Speed of car A = 36 km hr^-1
Converting speed of car A in m/sec by multiplying by 5/18:
= 36 x 5/18
= 6 x 5
= 30 m/sec
Speed of car B = 54 km hr^-1
Converting speed of car B in m/sec by multiplying by 5/18:
= 54 x 5/18
= 3 x 5
= 15 m/sec
To Find:
Minimum acceleration of car B required to avoid an accident
The speed of car B deducted from A and speed of car C added with A.
Calculating:
Calculating relative speed of Car A with respect to Car C
Vac = (10 + 15) ms^-1
= 25 ms^-1
Calculating relative speed of Car B with respect to Car A:
Vba = (15 - 10) ms^-1
= 5 ms^-1
Calculating the time taken by Car C to cover distance:
Using formula
t = 1000 / Vac
Substituting the values which we calculated into this formula we get:
t = 1000 / 25
t = 40 seconds
Now,
Calculating time taken by car C to cover distance AC
In this case if a is the acceleration:
We use the formula of motion:
s = ut + 1/2 at^2
Substituting values into this formula we get:
1000 = 5 x 40 + 1/2 a x 40 x 40
1000 = 200 + 800 a
1000 - 200= 800 a
800 = 800a
Taking a to the other 800 from the a to the other side we get:
a = 800 / 800
a = 1 m/s^2
Therefore, the minimum acceleration of car B required to avoid the accident is 1 m/s^2.
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