Question

After a rocket is launched at t = 0, its position is given by x(t)=qt^3, where q is some positive constant.

a. Sketch a graph of the rocketâs position as a function of time.
b. What is the rocketâs displacement during the interval from t = T to t = 3T?

Answer 1

Answer:

Explanation:

Consider that the position of the rocket launched is expressed by:

$x(t)=qt^3$

a)

As q is an arbitrary constant, therefore draw the lot between $\frac{x}{q}$ and $t^3$ as shown below.

The required plot is shown below in attachment

b)

The position of the rocket at time $t=T$ is

$x_T=qT^3$

The position of the rocket at time $t=2T$ is

$x_{3T}=q(3T)^3\\\\=27qT^3$

Then the displacement of the rocket is

$\bigtriangleup \vec x=(x_{3T}-x_T)\bar i\\\\=(27qT^3-qT^3)\bar i\\\\=26qT^3\bar i$

Therefore the displacement of the rocket between the times T and 3T is $\bigtriangleup \vec x=26qT^3\bar i$

Answer 2

The displacement is a vector quantity. The displacement of the rocket during the interval from T to 3T is $26qt^3 i$.

Displacement :

As given in the question, the position of the rocket launched is expressed by ,

$x(t) = qt^3$

Where,

$q$- arbitrary constant

$t$ - time

The position of the rocket at time T,

$x_T = qT^3$

The position of the rocket at time 3T,

$x_{3T }= q(3T)^3$

The displacement of the rocket,

$\Delta \vec x = 27qt^3- qT^3i\\\\\Delta \vec x = 26qt^3 i$

Therefore, the displacement of the rocket during the interval from T to 3T is $26qt^3 i$.

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