Q51E - University Physics with Modern Physics

Question

A rocket starts from rest and moves upward from the surface of the earth. For the first 10.0 s of its motion, the vertical acceleration of the rocket is given by a_y = (2.80 m/s³) t, where the +y-direction is upward. (a) What is the height of the rocket above the surface of the earth at t = 10.0 s? (b) What is the speed of the rocket when it is 325 m above the surface of the earth?

Step-by-Step Solution

Answer

(a)  Height of the rocket above the surface of the is 466.66 m.

(b)  Speed of the rocket is 109.89 m/s² 

 

Step 1: Identification of given values

$a\left(t\right)\mathit{=}2.80t$ 

Integrating above equation

$v\left(t\right)=1.4t^{2+}c$ 

Here c is the constant.

 

At t=0 , v=0 so that c=0

$v\left(t\right)=1.4t^2$ 

 

Again integrating

$x\left(\mathrm{t}\right)=\frac{1.4{\mathrm{t}}^3}{3}+\mathrm{d}$ 

Again d is the constant.

At t=0 , x(height)=0 so that d=0

$x\left(t\right)=\frac{1.4{\mathrm{t}}^3}{3}$

Now, t = 10 s

 $$\begin{gathered} x\left(t\right)=\frac{1.4\times {10}^3}{3} \\ x\left(t\right)=466.66 \mathrm{m} \end{gathered}$$ 

Step 2: calculation for the speed

$$\begin{gathered} 325=\frac{1.4\times t^3}{3} \\ t=8.86 \mathrm{s} \\ \mathbf{Now} \\ \mathrm{v}\left(\mathrm{t}\right)=1.4t^2 \\ v\left(t\right)=1.4\times \left(8.{16}^2\right) \\ v\left(t\right)=109.89 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$