Example 6-27. The acceleration of a particle is given by
a = sin tˆe 1 + cos tˆe 2.
If at time t= 0 the position and velocity of the particle are given by
r (0) = 6 ˆe 1 − 3 ˆe 2 + 4 ˆe 3 and v (0) = 7 ˆe 1 − 6 ˆe 2 − 5 ˆe 3 ,
find the position and velocity as a function of time.
Solution: An integration of the acceleration with respect to time produces the ve-
locity and ∫
a (t)dt =v =v (t) = −cos tˆe 1 + sin tˆe 2 +c 1 ,
where c 1 is a vector constant of integration. From the above initial condition for the
velocity, the constant c 1 can be determined. One finds
v (0) = −ˆe 1 +c 1 = 7 ˆe 1 − 6 ˆe 2 − 5 ˆe 3 or c 1 = 8 ˆe 1 − 6 ˆe 2 − 5 ˆe 3.
Consequently, the velocity can be expressed as a function of time in the form
v =v (t) = dr
dt
= (−cos t+ 8) ˆe 1 + (sin t−6)ˆe 2 − 5 ˆe 3.
An integration of the velocity with respect to time produces the position vector as
a function of time and
∫
v (t)dt =
∫
dr
dt
dt =
∫
(−cos t+ 8) dt ˆe 1 +
∫
(sin t−6) dt ˆe 2 − 5
∫
dt eˆ 3 +c 2
r (t) = (−sin t+ 8 t)ˆe 1 + (−cos t− 6 t)ˆe 2 − 5 tˆe 3 +c 2 ,
where c 2 is a vector constant of integration. From the above initial conditions, at
time t= 0 ,one can determine this vector constant of integration and
r (0) = −ˆe 2 +c 2 = 6 ˆe 1 − 3 ˆe 2 + 4 ˆe 3 or c 2 = 6 ˆe 1 − 2 ˆe 2 + 4 ˆe 3.
The position vector as a function of time can be expressed as
r =r (t) = (−sin t+ 8 t+ 6) ˆe 1 + (−cos t− 6 t−2)ˆe 2 + (− 5 t+ 4) ˆe 3.
