Similarly, the velocity of particle 2 is
2
22
0
2
20.0
( 8.00 )
20.0 4.00 .
v
v
a dt
t dt
t
=+
=
+
−
=
−
³³
The condition that
12
vv
=
implies
12.0
3.00
20.0 4.00
4.00
12.0
17.0
0
tt
t
t
+=−
¡
+−
=
which can be solved to give (taking positive root)
( 3
26)/ 2 1.05 s.
t
=−+
=
Thus, the
velocity at this time is
12.0(1.05) 3.00 15.6 m/s.
==
+
=
70. To solve this problem, we note that velocity is equal to the time derivative of a
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 Spring '06
 Stoler
 Physics, Calculus, Derivative, Acceleration, Velocity, dt dt

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