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Mathematics
Mathematics
Problem
The position of a particle is
r
(
t
)
=
⟨
t
3
,
3
t
2
−
6
t
,
4
⟩
\mathbf r(t)=\langle t^{3},\, 3t^{2}-6t,\,4\rangle
r
(
t
)
=
⟨
t
3
,
3
t
2
−
6
t
,
4
⟩
. For what value of
t
t
t
in the interval
0
≤
t
≤
2
0\le t\le 2
0
≤
t
≤
2
are the velocity and acceleration vectors perpendicular? a)
t
=
1
t=1
t
=
1
b)
t
≈
0.77
(
the real root of
t
3
+
2
t
−
2
=
0
)
t\approx0.77\;(\text{the real root of }t^{3}+2t-2=0)
t
≈
0.77
(
the real root of
t
3
+
2
t
−
2
=
0
)
c)
t
=
2
3
t=\sqrt{\dfrac{2}{3}}
t
=
3
2
d) There is no such
t
t
t
in the interval.
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