A spaceship maneuvering in space far from any gravitational

A spaceship maneuvering in space, far from any gravitational influences, is executing a predetermined acceleration program which yields a position vector r(t) for the ship, relative to a small space beacon, given by r(t) = (t - 2) i + (t - 3)^2 j + (t - 4)^3 k. a) Suppose that the captain shuts down the engines at time t_degree. Find the subsequent motion of the ship. b)Show that If t_degree is chosen appropriately then the ship will hit the beacon.

Solution

a)

v(t)=i+2(t-3)j+3(t-4)^2k

So if the engine shuts down the at t0 the sped is

v(t)=i+2(t0-3)j+3(t0-4)^2k

Acceleration is 0 so the ship moves with this uniform velocity

b)

Let the beacon be along the vector:ai+bj+ck

So, if we shut down at some time , to so that velocity vector is aligned at t0 with the beacon direction then we will reach the beacon

So, we need

i+2(t0-3)j+3(t0-4)^2k =k(ai+bj+ck)

ka=1

kb=2(t0-3)

t0=kb/2+3

kc=3(t0-4)^2=3(t0-3-1)^2=3(kb/2-1)^2

So we can solve for k. Hence an appropriate t0 can be chosen\'