Question

(a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t) = 20t i + sin(t) j + cos(2t) k, v(0) = i, r(0) = j

Answer 1

Answer:

$r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k$

Explanation:

a(t)=20t i+sin(t) j +cos(2t) k

v(t)=$\int\limits^a_b {a(t)} \, dt$

=$(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k$---------------eqn 1

given v(0)=i

i=$c_1i+(-1+c_2)j+(0+c_3)k$

$c_1=1$   $c_2=1$  $c_3=0$

from equation 1

V(t)=$(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k$----------eqn 2

now r(t)=$\int\limits^a_b {v (t)} \, dt$

$(\frac{10t^3}{3}+t+c_1)i+(-sint+t+c_2)j+(\frac{-cos2t}{4}+c_3)k$

given r(0)=j

0i+1j+ok = $c_1i+c_2j+(\frac{-1}{4}+c_3)k$

$c_1=0$  $c_2=0$  $c_3 = \frac{1}{4}$

$r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k$