Question

A 1000 kg weather rocket is launched straight up. The rocket

motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance. What was the rocket’s acceleration during the first 16 s?

Short answer

The rocket's acceleration during the first$16s$is$29.97m/s^2.$

Step-by-step solution

Given data

Time of rocket motion, $t=16s.$

The rocket altitude after time$t\text{'}=20s$ is, $s=5100m.$

We have to find the acceleration of the rocket.

Determination of displacement of the rocket in 16 s

Speed of the rocket after $16s,$

$$\begin{gathered} v=u+at \\ v=0+16a \\ v=16a...\left(1\right) \end{gathered}$$

Displacement of the rocket in $16s,$

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ s=0+\frac{1}{2}a{t}^2 \\ s=\frac{1}{2}a{t}^2...\left(2\right) \end{gathered}$$

Determination of displacement of the rocket  after 16 s

After moving for $16s,$the motor of the rocket stops and its altitude after $\left(20-16\right)s=4s$becomes $5100m.$

So the displacement of the rocket after $16s$is,

role="math" localid="1648488020296" $$\begin{gathered} x=vt+\frac{1}{2}g{t}^2 \\ x=16a\times 4-\frac{1}{2}\times 9.8\times 4^2 \\ x=64a-78.4...\left(3\right) \end{gathered}$$

Determination of the acceleration

According to the question, the altitude of the rocket after $16s$is, $x+s=5100$

Thus from equation $\left(2\right),\left(3\right)$, we can write,

role="math" localid="1648488126578" $$\begin{gathered} 5100=\left(\frac{1}{2}a\times {16}^2\right)+\left(64a-78.4\right) \\ 5100=128a+64a-78.4 \\ 192a=5100+78.4 \\ a=\frac{5178.4}{192} \\ a=29.97m/s^2 \end{gathered}$$