Question

A “sun yacht” is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of 900 kgand receives a push of 20 N.

(a) What is the magnitude of the resulting acceleration? If the craft starts from rest, (b) how far will it travel in 1 day and (c) how fast will it then be moving?

Short answer

(a) Magnitude of resulting acceleration is $0.022\mathrm{m}/{\mathrm{s}}^2$.

(b) Distance traveled by the craft in one day $8.21\times {10}^7\mathrm{m}$.

(c) Speed of the craft at the end of one day is $1.9\times {10}^3\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given data

• Mass of spacecraft $\left(m\right)\mathrm{is}900\mathrm{kg}$.
• The force on the spacecraft (F) is 20 N.

Understanding the concept of force and acceleration

Newton’s second law states that the net force$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$ on a body with mass mis related to the body’s acceleration $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}$by,

$\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}\mathbf{=}\mathit{m}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{a}}$

Given the mass and force of the spacecraft, we can calculate the acceleration of the spacecraft. In addition, by using the kinematic equation we can determine the distance covered by it in a certain time and find its speed.

Formulae:

$$\begin{gathered} F=ma \\ \left(\mathrm{i}\right) \\ {V}_f=V_0+at \\ \left(\mathrm{ii}\right) \\ s=v_{0}t+\frac{1}{2}a{t}^2 \\ \left(\mathrm{iii}\right) \\ 1\mathrm{day}=86400\mathrm{s} \\ \left(\mathrm{iv}\right) \end{gathered}$$

(a) Calculate the magnitude resulting acceleration

From equation (i), the acceleration can be written as,

$a=\frac{F}{m}$

Substitute the values in the acceleration equation.

$$\begin{gathered} a=\frac{F}{m} \\ =\frac{20\mathrm{N}}{900\mathrm{kg}} \\ =0.022\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the acceleration of the spacecraft is $0.022\mathrm{m}/{\mathrm{s}}^2$.

(b) Calculate the distance travelled by the craft in one day

Use the equation (iii) to calculate the distance traveled by the spacecraft in one day. We know the value of acceleration, time and initial velocity.

$$\begin{gathered} s=v_{0}t+\frac{1}{2}a{t}^2 \\ =0+\frac{1}{2}\left(0.022\mathrm{m}/{\mathrm{s}}^2\right)\left(86400s^2\right) \\ =82114560\mathrm{m} \\ =8.21\times {10}^7\mathrm{m} \end{gathered}$$

Therefore, the distance traveled by spacecraft in one day is $8.21\times {10}^7\mathrm{m}$.

(c) Calculate the speed of the craft

Use the equation (ii), acceleration and time value to calculate the speed of the spacecraft in one day,

$$\begin{gathered} v_f=0+at \\ =\left(0.022\mathrm{m}/{\mathrm{s}}^2\right)\left(86400\mathrm{s}\right) \\ =1900.8\mathrm{m}/\mathrm{s} \\ =1.9\times {10}^3 \end{gathered}$$

Therefore, the speed of the spacecraft at the end of one day is $1.9\times {10}^3$.