Space Walk

Say, David accelerates for $x$ seconds. After this acceleration, his velocity is $0.8x \text{ m/s}$ and the distance he covers while accelerating is $s_1 = 0.4x^2 \text{ m}$ . After that, David decelerates for $\dfrac{0.8x-4}{0.8}$ seconds, with an average velocity of $\dfrac{0.8x+4}{2} \text{ m/s}$ . The distance David covers while decelerating is $s_2 = \dfrac{0.8x-4}{0.8} \cdot \dfrac{0.8x+4}{2} = \dfrac{0.64x^2 - 16}{1.6} = 0.4x^2 - 10 \text{ m}.$ Hence, the total distance David travels is $s_1 + s_2 = 0.4x^2 + (0.4x^2 - 10) = 0.8x^2 - 10 = 310 \text{ m} \implies x = 20 \text{ seconds}.$ David decelerates for $\dfrac{0.8x-4}{0.8} = 15$ seconds, for a total time of $20 + 15 = \boxed{35}$ seconds.

Say instead of arriving at the satellite, David decelerates to $0 m/s$ .

The extra time he needs is $\frac{4m/s}{0.8m/s^2}=5s$ .

The extra distance he travels is $\frac{0.8m/s^2 \times (5s)^2}{2}=10m$ .

Hence his total distance is $310m+10m=320m$ .

Due to symmetry, we can now say he accelerates for $160m$ (which is half of $320m$ ).

Time spent accelerating $=\sqrt{\frac{2 \times 160m}{0.8m/s^2}}=20s$ .

He also spent $20s$ decelerating (due to symmetry).

There is $5s$ extra we added in front.

So, the time David spends $=20s+20s-5s=35s$ .

$s = s_1 + s_2$

$V^2_t = V^2_0 + 2 \cdot a \cdot s$

$s = \frac{V^2_t - V^2_0}{2 \cdot a}$

$V_t$ of $s_1$ = $V_0$ of $s_2$ let's call it as $V$

$s_1 = \frac{V^2 - 0}{1.6}$

$s_2 =\frac{V^2 - 4^2}{1.6}$

$s = s_1 = s_2$ $s=\frac{V^2}{1.6} + \frac{V^2-16}{1.6}$

$310 = \frac{2V^2-16}{1.6}$

$512 = 2V^2$ $256 = V^2$

$V = 16$ $V=V_0 + a \cdot t_1$ $16=0.8\cdot t_2$

$t_1$ = 20 $4=16-a\cdot t_2$

$t_2=15$ $t=t_1+t_2$

$t=35$

In order to solve this problem, I used the Work-Energy Theorem and a few Kinematics equations. We were given with:

$a = 0.8\\ v_1 = 0\\ v_3 = 4$

$v_1$ corresponds to his speed at time $t_1$ . Similarly, $v_3$ corresponds to his speed at time $t_3$ . We do not know (yet) what is $v_2$ or $t_2$ .

For $t_1$ to $t_2$ , the Work-Energy Theorem states that:

$\frac{1}{2}mv_2^{2} - \frac{1}{2}mv_1^{1} = ma\cdot{x}$

where $x$ is the displacement he has traveled for the first part of the 310-m trip. Cancelling $m$ and revising the equation we have:

$v_2^{2} - v_1^{2} = 2a\cdot{x}$

Now, for the second part of the trip, that is, from $t_2$ to $t_3$ :

$\frac{1}{2}mv_3^{2} - \frac{1}{2}mv_2^{2} = -ma\cdot{(310-x)}$

The total work done for this time interval is negative because he has slowed down. Again, cancelling $m$ and readjusting the equation:

$v_3^{2} - v_2^{2} = -2a\cdot{(310-x)}$

Adding equations 2 and 4, we will have:

$v_3^{2} - v_1^{2} = -2a\cdot{(310-2x)}$

But, since $v_1 = 0$ , we should get:

$\cfrac{v_3^{2}}{2a} + 310 = 2x$

By substitution, $x = 160$ . Going back to equation 2, we should get $v_2 = 16$ . Now, substituting all the values to the following, and thus getting $t_1$ and $t_2$ :

$a = \cfrac{v_2 - v_1}{t_1}\\ -a = \cfrac{v_3 - v_2}{t_2}\\ t_1 + t_2 = \boxed{35}$