On a two-lane road, car A is travelling with a speed of 36 km/hr. Two cars B and C approach car A in opposite directions with a speed of 54 km/hr 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 ?
Answers
Answer:
Explanation:
Solution,
Speed of Car A, v(A) = 36 km/h = 36 × 5/18 = 10 m/s.
Speed of Car B, v(B) = 54 km/h = 54 × 5/18 = 15 m/s.
Now,
Relative Speed of car A with respect to car C = v(AC) = 10 + 15 = 25 m/s
Relative Speed of car B with respect to car A = v(BA) = 15 - 10 = 5 m/s
Here, we know that
Time taken, t = Dsitance/Speed
⇒ t = 1000/v(AC)
⇒ t = 1000/25
⇒ t = 40 seconds
Here, the time taken is 40 seconds.
Now, Acceleration a
We know 2nd equation of motion,
s = ut + 1/2 × at²
Putting all the values, we get
⇒ 1000 = 5 × 40 + 1/2 × a × (40)²
⇒ 1000 = 200 + 1/2 × a × 1600
⇒ 1000 = 200 + 800a
⇒ a = 1000 - 200/800
⇒ a = 1 m/s².
Hence,The minimum required acceleration of car B to avoid an accident is 1 m/s².
Answer :
- Speed of A = 36km/h = 10m/s
- Speed of B = Speed of C = 54km/h = 15m/s
Relative speed of A w.r.t C= 10 + 15 = 25m/s
Time taken by C to overtake A (t) = 25/1000 = 40 s
Distance travelled by A= 10 * 40 = 400m
So, B have to cover distance of ( 1000m + 400m) = 1400 m to overtake A before C
So, we get :
- Distance (s) = 1400m
- Initial velocity (u) = 15 m/s
- Time = 40s
Substituting above values in 2nd Equation of Kinematics :
➝ s = ut + ½at²
➝ 1400 = 15 * 40 + ½ * a * (40)²
➝ 1400 = 600 + ½ * a * 1600
➝ 1400 - 600 = 800a
➝ 800 = 800a
➝ a = 800/800
➝ a = 1 m/s²
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