Q56P - University Physics with Modern Physics

Question

A lunar lander is descending toward the moon’s surface. Until the lander reaches the surface, its height above the surface of the moon is given by $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{b}\mathbf{-}\mathbf{ct}\mathbf{+}{\mathbf{dt}}^{\mathbf{2}}$, where $\mathbf{d}\mathbf{=}\mathbf{800}\mathbf{ }\mathbf{m}$ is the initial height of the lander above the surface, c=6.0 m/s , and d = 1.05 m/s². (a) What is the initial velocity of the lander, at t =0? (b) What is the velocity of the lander just before it reaches the lunar surface?

Step-by-Step Solution

Answer

a) the initial velocity of the lander is $-60 \frac{m}{s}$ and b) the velocity of the lander just before it reaches to the lunar surface is $-15.5\hspace{0.33em}\frac{m}{s}$.

Step 1: Identification of the given data

The given data can be expressed below as:

• The initial height of the lander is 800 m.
• The value of c and d are $60.0 \mathrm{m}/\mathrm{s} \mathrm{and} 1.05\mathrm{m}/{\mathrm{s}}^2$.

Step 2: Significance of the Newton’s first law for the lunar lander

This law illustrates that an object will continue to be in rest or in uniform motion unless it is resisted by an external force upon it.

Differentiating the equation of a motion along a straight line gives the initial velocity and using t =0, may be beneficial for the lunar lander.

Step 3: Determination of the initial velocity and the velocity of the lander

The given equation can be expressed as:

$y\left(t\right)=800\hspace{0.33em}m-(60.0\hspace{0.33em}m/s)t+(1.05\hspace{0.33em}m/s^2)t^2$

a) We know that, $v_y=\frac{dy}{dt}$

Hence, from the above equation, it can be expressed as:

$$\begin{gathered} v_y=\frac{d}{dt}\left(800-60t+1.05t^2\right) \\ =-60+2.1 t \end{gathered}$$

The initial velocity of the lander at t =0 , will be-

$$\begin{gathered} v_{y0}=-60+2.1\times 0 \\ =-60\frac{m}{s} \end{gathered}$$

Thus, the initial velocity of the lander is $-60\frac{m}{s}$.

b) the lunar lander reaches the surface when $y\left(t\right)=0$

$1.05{\mathrm{t}}^2-60\mathrm{t}+800=0$

By solving the quadratic equation, we can get two roots which are ${\mathrm{t}}_1=35.9\hspace{0.33em}\mathrm{s}\hspace{0.33em}\mathrm{and}\hspace{0.33em}{\mathrm{t}}_2=21.2\hspace{0.33em}\mathrm{s}$. Hence, from the gathered roots, we will take $t_2=21.2\hspace{0.33em}s$ as it is the least time to reach to the lunar surface.

The velocity of the lander before it reaches to the lunar surface is expressed as:

$$\begin{gathered} v_y=-60\frac{\mathrm{m}}{\mathrm{s}}+2\cdot 1\frac{\mathrm{m}}{{\mathrm{s}}^2}\cdot 21.2\mathrm{s} \\ =-15.5\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Thus, the velocity of the lander just before it reaches to the lunar surface is $-15.5\frac{\mathrm{m}}{\mathrm{s}}$.