Engineering Mechanics - Kinematics of Particle (KOP) - Discussion
Discussion Forum : 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.
Discussion:
1 comments Page 1 of 1.
Gaurav said:
1 decade ago
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)
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)
(1)
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