Engineering Mechanics - Kinematics of Particle (KOP) - Discussion

Discussion :: Kinematics of Particle (KOP) - General Questions (Q.No.2)

2. 

A two-stage missile is fired vertically from rest with an acceleration as shown in the graph. In 15 s the first stage A burns out and the second stage B ignites. How fast is the rocket moving and how far has it gone at t = 20 s? How fast is the missile moving and how far has it gone at t = 20 s?

[A]. v = 430 m/s, s = 4.30 km
[B]. v = 395 m/s, s = 3.69 km
[C]. v = 360 m/s, s = 3.60 km
[D]. v = 500 m/s, s = 5.00 km

Answer: Option B

Explanation:

No answer description available for this question.

Robin said: (May 26, 2011)  
Can anyone explain.

Abhilash Raj said: (Aug 20, 2012)  
Area=18*15=270m/s=initial velocity(U)
V=U+at
=270+25*5
=395m/s

s=d/t
395=d/20
d=7.9km

Rajesh Kumar said: (Apr 7, 2013)  
v = area of shaded region = 18*15+25*5 = 395m/s.

s = ut+1/2at*t = 2025m.

s1 = u1t1+1/2a1*t1*t1 = 1662.5m.

Total distance = s+s1 = 3687.5m = 3.69km.

Basky said: (Mar 12, 2014)  
v = Area of shaded region = 18*15+25*5 = 395m/s.
s = Average velocity x time.
= 395/2 x 20.
= 3950.
= 3.95m.

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