We define the origin $x=0$ at the starting point and let $t=0$ be the moment object B starts. Object A has a head start of $t_0 = 5$ s. The equations of motion for the positions of A and B are:

$x_A(t) = v_A(t + t_0)$ $x_B(t) = \frac{1}{2}a_B t^2$

The given values are $v_A = 1$ m/s, $t_0 = 5$ s, and $a_B = 40$ cm/s² = 0.4 m/s².

[Q1] How many seconds after B starts will it catch up to A? Catch-up occurs when $x_A(t) = x_B(t)$. Let this time be $t_{catch}$.

$$v_A(t_{catch} + t_0) = \frac{1}{2}a_B t_{catch}^2$$

Rearranging into a quadratic equation for $t_{catch}$:

$$\frac{1}{2}a_B t_{catch}^2 - v_A t_{catch} - v_A t_0 = 0$$

Using the quadratic formula and taking the positive root for time:

$$t_{catch} = \frac{v_A + \sqrt{v_A^2 - 4(\frac{1}{2}a_B)(-v_A t_0)}}{a_B} = \frac{v_A + \sqrt{v_A^2 + 2a_B v_A t_0}}{a_B}$$

Substituting the values:

$$t_{catch} = \frac{1 + \sqrt{1^2 + 2(0.4)(1)(5)}}{0.4} = \frac{1 + \sqrt{1+4}}{0.4} = \frac{1 + \sqrt{5}}{0.4} \approx 8.09 \text{ s}$$

[Q2] What is the maximum distance between them before they meet? The distance between the objects is $d(t) = x_A(t) - x_B(t)$. The distance is maximized when their velocities are equal, i.e., $v_B(t) = v_A$. The velocity of B is $v_B(t) = a_B t$. Let $t_{max}$ be the time when the distance is maximum.

$$a_B t_{max} = v_A \implies t_{max} = \frac{v_A}{a_B}$$

The maximum distance $d_{max}$ is the distance at this time:

$$d_{max} = d(t_{max}) = v_A(t_{max} + t_0) - \frac{1}{2}a_B t_{max}^2$$

Substituting $t_{max} = v_A/a_B$:

$$d_{max} = v_A\left(\frac{v_A}{a_B} + t_0\right) - \frac{1}{2}a_B\left(\frac{v_A}{a_B}\right)^2 = \frac{v_A^2}{a_B} + v_A t_0 - \frac{v_A^2}{2a_B}$$ $$d_{max} = \frac{v_A^2}{2a_B} + v_A t_0$$

Substituting the values:

$$d_{max} = \frac{1^2}{2(0.4)} + (1)(5) = 1.25 + 5 = 6.25 \text{ m}$$

[Q3] How far is the meeting point from the starting point? The meeting distance $x_{catch}$ is the position of either object at $t_{catch}$. Using the equation for B:

$$x_{catch} = x_B(t_{catch}) = \frac{1}{2}a_B t_{catch}^2$$

Substituting the expression for $t_{catch}$ from Q1:

$$x_{catch} = \frac{1}{2}a_B \left( \frac{v_A + \sqrt{v_A^2 + 2a_B v_A t_0}}{a_B} \right)^2 = \frac{(v_A + \sqrt{v_A^2 + 2a_B v_A t_0})^2}{2a_B}$$

Substituting the numerical value of $t_{catch}$:

$$x_{catch} = \frac{1}{2}(0.4)\left(\frac{1 + \sqrt{5}}{0.4}\right)^2 = 0.2 \frac{(1+\sqrt{5})^2}{0.16} = \frac{(1+\sqrt{5})^2}{0.8} = \frac{6+2\sqrt{5}}{0.8} = \frac{3+\sqrt{5}}{0.4} \approx 13.09 \text{ m}$$