Chapter 4, Problem 42RE

To determine

To find : How fast will rocket is going 1 min later.

Answer to Problem 42RE

The answer: $1200\text{m}/\text{sec}$

Explanation of Solution

Given information:

The rocket lifts of the surface of earth with a constant acceleration $20\text{m}/{\text{sec}}^2$ .

Calculation:

  $a(t)=20\left(\text{m}/{\text{sec}}^2\right)$

Velocity is the anti derivative of the acceleration.

So,

  $v(t)=20t+C\mathrm{.....}(1)$

Hence, a rocket lift off $\Rightarrow v(0)=0$

Put, $t=0$ into equation $(1)$ and obtain:

  $\begin{array}{c}0=20\cdot 0+C\\ \Rightarrow C=0\\ v(t)=20t(\text{m}/\text{sec})\end{array}$

Convert the given time, 1 minute, into seconds because the units of velocity are meters per second. 60 seconds in a minute.

  $\begin{array}{c}v(60)=20\cdot 60\\ =1200(\text{m}/\text{sec})\end{array}$

Therefore, the required $1200\text{m}/\text{sec}$ will rocket is going 1 min later.