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Mathematics
Problem
A particle moves along a line with velocity
v
(
t
)
=
k
t
2
−
6
t
+
4
v(t)=kt^{2}-6t+4
v
(
t
)
=
k
t
2
−
6
t
+
4
for
t
≥
0
t\ge 0
t
≥
0
. The particle is at the origin at both
t
=
0
t=0
t
=
0
and
t
=
4
t=4
t
=
4
. What is its position at
t
=
1
t=1
t
=
1
? a)
17
12
\dfrac{17}{12}
12
17
b)
3
2
\dfrac32
2
3
c)
2
2
2
d)
7
6
\dfrac76
6
7
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