Engineering Mechanics - Kinematics of Particle (KOP) - Discussion

Discussion :: Kinematics of Particle (KOP) - General Questions (Q.No.9)

9. 

The pilot of flighter plane F is following 1.5 km behind the pilot of bomber B. Both planes are originally traveling at 120 m/s. In an effort to pass the bomber, the pilot in F gives his plane a constant acceleration of 12 m/s2. Determine the speed at which the pilot in the bomber sees the pilot of the fighter plane pass at the start of the passing operation the bomber is decelerating at 3 m/s2. Neglect the effect of any turning.

[A]. vF/B = 150 m/s
[B]. vF/B = 367 m/s
[C]. vF/B = 90 m/s
[D]. vF/B = 212 m/s

Answer: Option D

Explanation:

No answer description available for this question.

Gaurav said: (Aug 22, 2011)  
Solution : Here,

Uf = Ub = 120 m/s (initial velocity)

Relative initial velocity = Urel = Uf - Ub = 120 - 120 = 0 m/s

Now,

Relative acceleration(at the start of passing), Arel = 12 - (-3) = 15 m/s2

Now,

Vrel^2 = Urel^2 + 2*Arel*s

where,

S = Distance between the aeroplanes = 1500 m

We have,

Vrel^2 = 0^2 + 2*15*1500 = 45000 m2/s2

Vrel = (4500)^(0.5) = 212 m/s (Answer)

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