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Engineering Mechanics - Kinematics of Particle (KOP) - Discussion

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

v = 430 m/s, s = 4.30 km
v = 395 m/s, s = 3.69 km
v = 360 m/s, s = 3.60 km
v = 500 m/s, s = 5.00 km
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
4 comments Page 1 of 1.
Basky said:   1 decade ago
v = Area of shaded region = 18*15+25*5 = 395m/s.
s = Average velocity x time.
= 395/2 x 20.
= 3950.
= 3.95m.
(2)
RAJESH KUMAR said:   1 decade ago
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.
(1)
Abhilash raj said:   1 decade ago
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
Robin said:   1 decade ago
Can anyone explain.
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