STEM Problem Detail

Problem Statement

Prompt

Give me full step by step solution of Question 12-61

Solution Rendering...

## Solution 1: Kinematics of Ball A and B

**Kinematics:** First, we consider the motion of ball $A$ with $(v_A)_0 = v_0$, $(s_A)_0 = 0$, $s_A = h$, $t_A = t'$, and $(a_c)_A = -g$.

$$(\text{+}{\uparrow}) \quad s_A = (s_A)_0 + (v_A)_0 t_A + \frac{1}{2}(a_c)_A t_A^2$$

$$h = 0 + v_0 t' + \frac{1}{2}(-g)(t')^2$$

$$h = v_0 t' - \frac{g}{2} t'^2 \tag{1}$$

$$(\text{+}{\uparrow}) \quad v_A = (v_A)_0 + (a_c)_A t_A = v_0 - gt' \tag{2}$$

The motion of ball $B$ requires $(v_B)_0 = v_0$, $(s_B)_0 = 0$, $s_B = h$, $t_B = t' - t$, and $(a_c)_B = -g$.

$$(\text{+}{\uparrow}) \quad s_B = (s_B)_0 + (v_B)_0 t_B + \frac{1}{2}(a_c)_B t_B^2$$

$$h = 0 + v_0(t' - t) + \frac{1}{2}(-g)(t' - t)^2$$

$$h = v_0(t' - t) - \frac{g}{2}(t' - t)^2 \tag{3}$$

$$(\text{+}{\uparrow}) \quad v_B = (v_B)_0 + (a_c)_B t_B = v_0 - g(t' - t) \tag{4}$$

Solving Eqs. (1) and (3):

**Ans.** $$v_0 t' - \frac{g}{2} t'^2 = v_0(t' - t) - \frac{g}{2}(t' - t)^2$$

$$\boxed{t' = \frac{2v_0 + gt}{2g}} $$

Substituting into Eq. (2):

**Ans** $$v_A = v_0 - g\!\left(\frac{2v_0 + gt}{2g}\right) = -\frac{1}{2}gt \quad \Rightarrow \quad v_A = \tfrac{1}{2}gt \downarrow $$

Substituting into Eq. (4):

**Ans** $$v_B = v_0 - g\!\left(\frac{2v_0 + gt}{2g} - t\right) = \frac{1}{2}gt \uparrow $$

---

## Solution 2: Escape Velocity Calculation

$$v\,dv = a\,dy$$

$$\int_{v}^{0} v\,dv = -g_0 R^2 \int_{0}^{\infty} \frac{dy}{(R + y)^2}$$

$$\left.\frac{v^2}{2}\right|_{v}^{0} = \left.\frac{g_0 R^2}{R + y}\right|_{0}^{\infty}$$

$$v = \sqrt{2g_0 R} = \sqrt{2(9.81)(6356 \times 10^3)} = 11{,}167\ \text{m/s} = 11.2\ \text{km/s}$$

---

## Solution 3: Variable Acceleration Integration

From Prob. 12-36:

$$(\text{+}{\uparrow}) \quad a = -g_0\frac{R^2}{(R + y)^2}$$

Since $a\,dy = v\,dv$:

$$-g_0 R^2 \int_{y_0}^{y} \frac{dy}{(R + y)^2} = \int_{0}^{v} v\,dv$$

$$g_0 R^2 \left[\frac{1}{R + y}\right]_{y_0}^{y} = \frac{v^2}{2}$$

$$g_0 R^2 \left[\frac{1}{R + y} - \frac{1}{R + y_0}\right] = \frac{v^2}{2}$$

Thus:

$$v = -R\sqrt{\frac{2g_0(y_0 - y)}{(R + y)(R + y_0)}}$$

When $y_0 = 500\ \text{km}$, $y = 0$:

$$v = -6356 \times 10^3 \sqrt{\frac{2(9.81)(500 \times 10^3)}{6356(6356 + 500) \times 10^6}}$$

$$v = -3016\ \text{m/s} = 3.02\ \text{km/s} \downarrow$$

---

## Solution 4: Total Distance Traveled by Train

**Total Distance Traveled:** The distance for part one of the motion is related to time $t = t'$ using $s_0 = 0$, $v_0 = 0$:

$$(\xrightarrow{+}) \quad s_1 = 0 + 0 + \frac{1}{2}(0.5)(t')^2 = 0.25(t')^2$$

The velocity at time $t'$:

$$(\xrightarrow{+}) \quad v = v_0 + a_c t = 0 + 0.5t' = 0.5t' \tag{1}$$

The time for the second stage is $t_2 = 160 - t'$, with constant velocity $v = 0.5t'$:

$$(\xrightarrow{+}) \quad s_2 = v\,t_2 = 0.5t'(160 - t') = 80t' - 0.5(t')^2$$

Setting $s_{\text{Tot}} = 2000\ \text{m}$:

$$s_{\text{Tot}} = s_1 + s_2$$

$$2000 = 0.25(t')^2 + 80t' - 0.5(t')^2$$

$$0.25(t')^2 - 80t' + 2000 = 0$$

Choosing the root less than 160 s:

$$t' = 27.3\ \text{s} $$ **Ans**

**$v$–$t$ Graph:** From Eq. (1), when $t = t' = 27.34\ \text{s}$: $v = 0.5(27.34) = 13.7\ \text{ft/s}$.

---

## Solution 5: Motion of Rockets A and B

**Rocket $A$**

For $t \leq 12\ \text{s}$:

$$+{\uparrow} \quad v_A = 0 + 20t = 20t$$

$$+{\uparrow} \quad s_A = 0 + 0 + \frac{1}{2}(20)t^2 = 10t^2$$

At $t = 12\ \text{s}$: $v_A = 240\ \text{m/s}$, $s_A = 1440\ \text{m}$.

For $t > 12\ \text{s}$:

$$v_A = 240\ \text{m/s}, \qquad s_A = 1440 + 240(t - 12)$$

**Rocket $B$**

For $t \leq 10\ \text{s}$:

$$+{\uparrow} \quad v_B = 0 + 15t = 15t$$

$$+{\uparrow} \quad s_B = 0 + 0 + \frac{1}{2}(15)t^2 = 7.5t^2$$

At $t = 10\ \text{s}$: $v_B = 150\ \text{m/s}$, $s_B = 750\ \text{m}$.

For $t > 10\ \text{s}$:

$$v_B = 150\ \text{m/s}, \qquad s_B = 750 + 150(t - 10)$$

At $t = 20\ \text{s}$: $s_A = 3360\ \text{m}$, $s_B = 2250\ \text{m}$, so $\Delta s = 1110\ \text{m} = 1.11\ \text{km}$.

---

## Solution 6: $a$–$t$ and $v$–$t$ Analysis

**$a$–$t$ Graph:**

$$a = \frac{dv}{dt} = \frac{d}{dt}(t^2 - 4t + 3) = (2t - 4)\ \text{m/s}^2$$

$$a\big|_{t=0} = -4\ \text{m/s}^2, \qquad a\big|_{t=2} = 0, \qquad a\big|_{t=4\ \text{s}} = 4\ \text{m/s}^2$$

**$v$–$t$ Graph:** The slope of the $v$–$t$ graph is zero when $a = \dfrac{dv}{dt} = 0$:

$$2t - 4 = 0 \implies t = 2\ \text{s}$$

Velocities at $t = 0,\ 2,\ 4\ \text{s}$:

$$v\big|_{t=0} = 3\ \text{m/s}, \qquad v\big|_{t=2\ \text{s}} = -1\ \text{m/s}, \qquad v\big|_{t=4\ \text{s}} = 3\ \text{m/s}$$

---

## Solution 7: Constant Acceleration Phase

$$v_1 = 0$$

$$v_2 = 162\ \frac{\text{mi}}{\text{h}} \times \frac{5280\ \text{ft}}{3600\ \text{s}} = 237.6\ \text{ft/s}$$

$$v_2^2 = v_1^2 + 2a_c(s_2 - s_1) \implies (237.6)^2 = 0 + 2\,a_c(5000) \implies a_c = 5.645\ \text{ft/s}^2$$

$$v_2 = v_1 + a_c t \implies 237.6 = 5.645\,t \implies t = 42.1\ \text{s}$$

$$v_3 = 220\ \frac{\text{mi}}{\text{h}} \times \frac{5280\ \text{ft}}{3600\ \text{s}} = 322.67\ \text{ft/s}$$

$$v_3^2 = v_2^2 + 2a_c(s_3 - s_2) \implies (322.67)^2 = (237.6)^2 + 2(3)(s - 5000) \implies s = 12{,}943\ \text{ft}$$

$$v_3 = v_2 + a_c t \implies 322.67 = 237.6 + 3\,t \implies t = 28.4\ \text{s}$$

---

## Solution 8: Kinematics of Package and Elevator

**Package:**

$$(\text{+}{\uparrow}) \quad v^2 = v_0^2 + 2a_c(s - s_0) = (4)^2 + 2(-32.2)(0 - 100)$$

$$v = 80.35\ \text{ft/s} \downarrow$$

$$(\text{+}{\uparrow}) \quad v = v_0 + a_c t \implies -80.35 = 4 + (-32.2)t \implies t = 2.620\ \text{s}$$

**Elevator:**

$$(\text{+}{\uparrow}) \quad s = s_0 + v\,t = 100 + 4(2.620) = 110\ \text{ft}$$

---

## Solution 9: Comparative Motion of Car A and Car B

**Car A** (for $0 \leq t \leq 10\ \text{s}$):

$$v_A = 0 + 4t = 4t, \qquad s_A = \frac{1}{2}(4)t^2 = 2t^2$$

At $t = 10\ \text{s}$: $v_A = 40\ \text{m/s}$, $s_A = 200\ \text{m}$.

For $t > 10\ \text{s}$ (constant velocity):

$$\int_{200}^{s_A} ds = \int_{10}^{t} 40\,dt \implies s_A = 40t - 200$$

At $t = 15\ \text{s}$: $s_A = 400\ \text{m}$.

**Car B** (for $0 \leq t \leq 5\ \text{s}$):

$$v_B = 5t, \qquad s_B = \frac{1}{2}(5)t^2 = 2.5t^2$$

When $v_B = 25\ \text{m/s}$: $t = 5\ \text{s}$, $s_B = 62.5\ \text{m}$.

At $t = 15\ \text{s}$: $s_B = 312.5\ \text{m}$.

Loading solution...

Problem #29