91Ó°ÊÓ

Consider the given question,

$v\left(t\right)=0.012t^2+400$

As the spaceship start with an initial velocity. Then,

$v_0=400$thousand miles per hour

The initial distance travel by the spaceship is $s_0=0$miles.

While the acceleration of the object will be $a_0=0$

Using anti derivative, the position function will be given as,

localid="1648496579742" $s\left(t\right)=0.004t^3+400t......\left(i\right)$

Using derivative, the acceleration will be given as,

$$\begin{gathered} d\left(t\right)=\frac{d}{dx}\left(0.012t^2+400\right) \\ d\left(t\right)=0.012\left(2t\right) \\ d\left(t\right)=0.024t \end{gathered}$$

As the acceleration of the object here is always positive. Therefore, it can be said that the spaceship always moves towards the sun.

Consider the acceleration function, it is observed that the acceleration of the object increases as the value of t increases.

Therefore, it can be said that the speed of the spaceship is accelerating.

Substitute $s\left(t\right)=67000$in equation (i),

role="math" localid="1648496818081" $$\begin{gathered} 0.004t^3+400t=67000 \\ 0.004t^3+400t-67000=0 \\ t≈140 \end{gathered}$$

Therefore, the time required is approximately $140$hrs or $6$days.