the position of a particle is given by x=t^3-6t^2+9t. find its position at which velocity becomes zero
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Answered by
0
Answer:
A
the body comes to rest firstly at (3−
7
) and then at (3+
7
)
C
the total displacement of the particle in travelling from the first zero of velocity to the second zero of velocity is −74 cm.
D
the particle reverses its velocity at (3−
7
) s and then at (3+
7
) s and has a negative velocity for (3−
7
) < t < (3+
7
)
x
t
=t
3
−9t
2
+6t
V=
dt
d
X
t
=(3t
2
−18t+6)cms
−1
For body to be at rest v=0
⇒t=(3±
7
)s
⇒t
1
=(3−
7
)s and ⇒t
2
=(3+
7
)s
Displacement of particle between t
1
and t
2
is x
t
2
−x
t
1
((3+
7
)
3
−9(3+
7
)
2
+6(3+
7
))−((3−
7
)
3
−9(3−
7
)
2
+6(3−
7
))=−74
v<0for(3−
7
)<t<(3+
7
)
Answered by
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Answer:
4 units
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