The velocity vector of the motion of particle,
described by the position vector P = 2t2q+t
is
(1) v = 4ti +3+2
(2) V = 4 +6t)
(3) v = 2t2 i +372)
(4)v=2i+j
Answers
Answer:
In parts (b) and (c), we use Eq. 4-10 and Eq. 4-16. For part (d), we find the direction
of the velocity computed in part (b), since that represents the asked-for tangent line. (a) Plugging into the given expression, we obtain
r∣
t=2.00
=[2.00(8)−5.00(2)]
i
^
+[6.00−7.00(16)]
j
^
=(6.00
i
^
−106
j
^
)m
(b) Taking the derivative of the given expression produces
v
(t)=(6.00t
2
−5.00)
i
^
−28.0t
3
j
^
where we have written v(t) to emphasize its dependence on time. This becomes, at t=2.00s,
v
=(19.0
i
^
−224
j
^
)m/s.
(c) Differentiating the
v(t)
found above, with respect to t produces 12.0t
i
^
−84.0t
2
j
^
, which yields
a
=(24.0
i
^
−336
j
^
)m/s
2
att=2.00s.
(d) The angle of
v
, measured from +x, is either
tan−1(
19.0m/s
−224m/s
)=−85.2
o
or94.8
o
where we settle on the first choice (–85.2°, which is equivalent to 275° measured counterclockwise from the +x axis) since the signs of its components imply that it is in
the fourth quadrant.
Explanation: