hopefully this is relatively simple and it's just me who doesn't know how to work it out. I tried integrating (finding the anti-derivative) the equation to find the velocity, then integrating again to find displacement but my answers aren't right.

"A rocket is launched from its pad at ground level and moves vertically upwards with its engines causing it to accelerate at (28-0.2t) m/s/s for the first two minutes, t being the time, in seconds, since the launch.

At the end of the two minutes the engines reduce the thrust they produce to a level just sufficient to maintain the vertical velocity achieved at that time.

How high is the rocket after a) 1 minute? b) 2 minutes? c) 3 minutes?"

I have the answers in the back of the book which are: 43.2km, 144km, and 259.2km. Could someone please show me step by step how to achieve " answers. Thank you.


We have

\begin{align} a(t) &= 28-0.2t \\ v(t) & = \int a(t) ~dt \\ & = \int 28 - 0.2t ~ dt \\ & = \frac{28t^1}{1} - \frac{0.2t^2}{2} + c_v \\ & = 28t - 0.1t^2 \end{align}

$c_v=0$ because initial velocity is $0$. (The rocket is resting on the ground). Let's continue to find $s(t)$.

\begin{align} s(t) &= \int v(t) ~ dt \\ & = \int 28t-0.1t^2 ~ dt \\ & = \frac{28t^2}{2} - \frac{0.1t^3}{3} + c_s \\ & = 14t^2 - \frac{0.1}{3}t^3 \end{align}

$c_s = 0$ because initial displacement is $0$. Now $1$ min $=$ $60$ sec.

\begin{align} s(60) &= 14(60)^2-\frac{0.1}{3}(60)^3 \\ & = 43200~m \\ & = 43.2~km \end{align}

Using a similar technique we find that $s(120) = 144 ~ km$ and $s(180) = 259.2 ~ km$. Hopefully this answers your question and you see where you went wrong.

  • 1
    $\begingroup$ You should say something about the fact that the rocket stops accelerating after $120~\text{s}$ since we cannot simply integrate the given velocity function from $t = 0$ to $t = 180$. $\endgroup$ – N. F. Taussig Mar 15 '18 at 11:36
  • $\begingroup$ @N.F.Taussig Yes, I did think of that. However, it does match up with the answer in the textbook. Perhaps it is a typo? $\endgroup$ – Landuros Mar 15 '18 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.