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

`r(t)=((10t^3)/(3)+t)i+(-sint+t+1)j+((-cos2t)/(4)+(1)/(4))k`

Step-by-step 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+(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+(sin2t)/(2)k`----------eqn 2

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

`((10t^3)/(3)+t+c_1)i+(-sint+t+c_2)j+((-cos2t)/(4)+c_3)k`

given r(0)=j

0i+1j+ok =
`c_1i+c_2j+((-1)/(4)+c_3)k`

`c_1=0`
`c_2=0`
`c_3 =  (1)/(4)`

`r(t)=((10t^3)/(3)+t)i+(-sint+t+1)j+((-cos2t)/(4)+(1)/(4))k`