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?

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

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:

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

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 $0.022\mathrm{m}/{\mathrm{s}}^2$.

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 $8.21\times {10}^7\mathrm{m}$.

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 $1.9\times {10}^3$.