## Suppose a rocket ship in deep space moves with constant acceleration equal to 9.8 m/s^2, which gives the illusion of normal gravity during the flight. a. If it starts from rest, how long will it take to acquire a speed one-tenth that of light, which travels at 3.0 x 10^8 m/s? b. How far will it travel in so doing?

$85.034 \setminus \setminus \textrm{h r s}$

$4.588164 \setminus \times {10}^{13} \setminus m$

#### Explanation:

The time required to achieve a velocity $v = \frac{1}{10} \setminus \cdot 3 \setminus \times {10}^{8} = 3 \setminus \times {10}^{7} \setminus \frac{m}{s}$ from rest $u = 0$ under an acceleration $a = 9.8 \setminus \frac{m}{s} ^ 2$

$v = u + a t$

$3 \setminus \times {10}^{7} = 0 + 9.8 t$

$t = 3.06 \setminus \times {10}^{6} \setminus s$

$t = 85.034 \setminus \setminus \textrm{h r s}$

The distance $s$ traveled by the rocket ship starting from rest $u = 0$ in time $t = 3.06 \setminus \times {10}^{6} \setminus s$ under an acceleration $a = 9.8 \setminus \frac{m}{s} ^ 2$ is given as follows

$s = u t + \frac{1}{2} a {t}^{2}$

$s = \left(0\right) \left(3.06 \setminus \times {10}^{6}\right) + \frac{1}{2} 9.8 {\left(3.06 \setminus \times {10}^{6}\right)}^{2}$

$= 4.588164 \setminus \times {10}^{13} \setminus m$