Question

What’s the velocity of the rocket? Use .

Answer 1

Answer:

25 m/s

Explanation:

Answer 2

Explanation:

Consider the given function  $K(t)=\frac{1}{2} m(t) \cdot v(t)^{2}$

Given that the velocity of rocket is increases at the rate of $15 m / \sec ^{2} \text { Hence } \frac{d v}{d t}=15$

The mass of rocket is decreasing at rate of $10 \mathrm{kg} / \mathrm{sec}$.

Hence $\frac{d m(t)}{d t}=-10$

Now find the rate of change of kinetic energy when $m(t)=2000 \mathrm{kg}=2 \times 10^{3} \mathrm{kg}$and $v(t)=5000 m / \mathrm{sec}^{2}=5 \times 10^{3} \mathrm{m} / \mathrm{sec}^{2}$

Now consider the given equation,

$\begin{aligned}&K(t)=\frac{1}{2} m(t) \cdot v(t)^{2}\\&\frac{d K}{d t}=\frac{1}{2}\left[m(t) \times 2 v(t) \times \frac{d v(t)}{d t}+v(t)^{2} \frac{d m}{d t}\right]\end{aligned}$

Now substituting the values $m(t)=2000 k g=2 \times 10^{3} k g$ and $v(t)=5000 m / \mathrm{sec}^{2}=5 \times 10^{3} \mathrm{m} / \mathrm{sec}^{2}$

$\begin{aligned}\frac{d K}{d t} &=\frac{1}{2}\left[2 \times 10^{3} \times 2 \times 5 \times 10^{3} \times 15+\left(5 \times 10^{3}\right)^{2}(-10)\right] \\&=\frac{1}{2}\left[300 \times 10^{6}+(-250) \times 10^{6}\right] \\&=\frac{1}{2}\left[(50) \times 10^{6}\right] \\&=25 M J\end{aligned}$

The rate at which kinetic energy changing is 25MJ