x=a/b(1+e^bt) find [a] and [b]
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Answer:
come back to its starting point at t→∞
x=
b
a
⎝
⎛
1−e
−bx
b
1
⎠
⎞
=
b
a
(1−e
−1
)
=
b
a
(1−
e
1
)
=
b
a
e
(e−1)
=
b
a
2.718
(2.718−1)
=
b
a
2.718
(1.718)
=0.637
b
a
=
3
2
b
a
velocity v=
dt
dx
=ae
−bt
,v
0
=a
acceleration a=
dt
dv
=−abe
−bt
&a
0
=−ab
At t=0,x=
b
a
(1−1)=0 and
At t=
b
1
,x=
b
a
(1−e
−1
)=
b
a
(1−
e
1
)=
3
2
b
a
At t=∞,x=
b
a
It cannot go beyond this, so point x>
b
a
is not reached by the particle.
At t=0,x=0, at t=∞,x=
b
a
, therefore the particle does not come back to its starting point at t=∞
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