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Mathematics
Problem
The displacement of a particle is
s
(
t
)
=
k
(
1
+
b
t
)
2
s(t)=\dfrac{k}{(1+bt)^{2}}
s
(
t
)
=
(
1
+
b
t
)
2
k
m, where
k
k
k
and
b
b
b
are positive constants. Which expression gives the velocity
v
(
t
)
v(t)
v
(
t
)
? a)
v
(
t
)
=
+
2
k
b
(
1
+
b
t
)
−
3
v(t)=+2kb(1+bt)^{-3}
v
(
t
)
=
+
2
k
b
(
1
+
b
t
)
−
3
b)
v
(
t
)
=
−
2
k
(
1
+
b
t
)
−
3
v(t)=-2k(1+bt)^{-3}
v
(
t
)
=
−
2
k
(
1
+
b
t
)
−
3
c)
v
(
t
)
=
−
2
k
b
(
1
+
b
t
)
−
3
v(t)=-2kb(1+bt)^{-3}
v
(
t
)
=
−
2
k
b
(
1
+
b
t
)
−
3
d)
v
(
t
)
=
−
2
k
b
(
1
+
b
t
)
−
2
v(t)=-2kb(1+bt)^{-2}
v
(
t
)
=
−
2
k
b
(
1
+
b
t
)
−
2
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