RC AGI BenchMark

The Strong Problems - For Advanced CalTech Students

Problem 1/167:

A Caltech freshman, inexperienced with suburban traffic officers, has just
received a ticket for speeding. - Thereafter, when he comes upon one of the "Speedometer Test" sections on a level stretch of highway, he decides to check his speedometer reading. As he passes the "0" start of the marked section, he presses his accelerator to the floor and for the entire period of the test he holds his car at constant maximum acceleration. He notices that he passes the 0.10 mile post 16 sec. after starting the test, and 8.0 sec. later he passes the 0.20 mile post.
(a) What should his speedometer have read at the 0. 2 mile post?
(b) What was his maximum acceleration?

- The car accelerates at a constant rate \( a \) after passing the 0-mile marker.
- Initial velocity at \( t = 0 \) is \( v_0 \).
- At \( t_1 = 16 \) seconds, the car is at \( s_1 = 0.10 \) miles.
- At \( t_2 = 24 \) seconds, the car is at \( s_2 = 0.20 \) miles.

1. **Write the equations of motion for \( s_1 \) and \( s_2 \):**
$$s_1 = v_0 t_1 + \frac{1}{2} a t_1^2 = 0.10 \text{ miles}$$
$$s_2 = v_0 t_2 + \frac{1}{2} a t_2^2 = 0.20 \text{ miles}$$

2. **Subtract the first equation from the second to eliminate \( v_0 \):**
$$s_2 - s_1 = v_0 (t_2 - t_1) + \frac{1}{2} a (t_2^2 - t_1^2)$$
$$0.10 = v_0 (8) + \frac{1}{2} a (320)$$
$$0.10 = 8v_0 + 160a \quad \text{(Equation 1)}$$

3. **Express \( s_1 \) in terms of \( v_0 \) and \( a \):**
$$0.10 = 16v_0 + 128a \quad \text{(Equation 2)}$$

4. **Subtract Equation 2 from Equation 1 to solve for \( v_0 \):**
$$0 = -8v_0 + 32a$$
$$v_0 = 4a$$

5. **Substitute \( v_0 \) back into Equation 2:**
$$0.10 = 16(4a) + 128a$$
$$0.10 = 192a$$
$$a = \frac{0.10}{192} \text{ miles/sec}^2$$

6. **Calculate \( v_0 \):**
$$v_0 = 4a = \frac{4 \times 0.10}{192} = \frac{0.10}{48} \text{ miles/sec}$$

7. **Find the velocity at \( t_2 = 24 \) seconds:**
$$v = v_0 + a t_2 = \frac{0.10}{48} + \left( \frac{0.10}{192} \times 24 \right)$$
$$v = \frac{0.10}{48} + \frac{0.10 \times 24}{192}$$
$$v = \frac{0.10}{48} + \frac{0.10}{8}$$
$$v = \frac{0.10}{48} + \frac{0.10}{8} = \frac{0.10(1 + 6)}{48} = \frac{0.10 \times 7}{48}$$

8. **Convert \( v \) to miles/hour:**
$$v = \frac{0.10 \times 7}{48} \times 3600 \text{ miles/hour}$$
$$v = \frac{0.70}{48} \times 3600 \approx 52.5 \text{ miles/hour}$$

$$a = \frac{0.10}{192} \text{ miles/sec}^2$$

2. **Convert miles to feet (\( 1 \text{ mile} = 5280 \text{ feet} \)):**
$$a = \frac{0.10 \times 5280}{192} \text{ ft/sec}^2$$
$$a = \frac{528}{192} \text{ ft/sec}^2 \approx 2.75 \text{ ft/sec}^2$$

Problem 2/167:

A Corporal rocket fired vertically is observed to have a constant upward acceleration of 2 g during the burning of the rocket motor, which lasted for 50 seconds. Neglecting air resistance and variation of g with altitude,
(a) draw V-t diagram for entire flight of rocket;
(b) calculate the maximum height attained;
(c) calculate the total elapsed time from the firing of the rocket to its return to earth.

- **Acceleration:** The rocket accelerates upward at \( a = +2g \) (where \( g = 9.8 \, \text{m/s}^2 \)).
- **Velocity Increase:** The velocity increases linearly from 0 to its maximum value at the end of this phase.[/markdown]
- **Graph Representation:** A straight line starting from the origin (0,0) and increasing with a slope of \( +2g \) up to \( t = 50 \, \text{s} \).
- **Acceleration:** After fuel burnout, the only acceleration is due to gravity, \( a = -g \).
- **Velocity Decrease:** The velocity decreases linearly from its maximum value down to zero at the peak of the trajectory.[/markdown]
- **Graph Representation:** A straight line with a slope of \( -g \) starting from the velocity at \( t = 50 \, \text{s} \) and decreasing to zero at \( t = 150 \, \text{s} \).
- **Graph Representation:** A straight line continuing from zero velocity at \( t = 150 \, \text{s} \) and decreasing (becoming more negative) until the rocket reaches the ground at approximately \( t = 272.5 \, \text{s} \).

- **At \( t = 0 \, \text{s} \):** \( v = 0 \, \text{m/s} \)
- **At \( t = 50 \, \text{s} \):** \( v = v_{\text{burnout}} = 980 \, \text{m/s} \)
- **At \( t = 150 \, \text{s} \):** \( v = 0 \, \text{m/s} \) (maximum height)
- **At \( t \approx 272.5 \, \text{s} \):** \( v = -v_{\text{impact}} \) (velocity upon returning to Earth)

- Acceleration during powered ascent: \( a_1 = +2g = +19.6 \, \text{m/s}^2 \)
- Time of powered ascent: \( t_1 = 50 \, \text{s} \)
- Acceleration during coasting ascent and descent: \( a_2 = -g = -9.8 \, \text{m/s}^2 \)
- Initial velocity: \( v_0 = 0 \, \text{m/s} \)

(\( v_{\text{burnout}} \))

\[
v_{\text{burnout}} = v_0 + a_1 t_1 = 0 + (19.6 \, \text{m/s}^2)(50 \, \text{s}) = 980 \, \text{m/s}
\]

(\( h_1 \))

\[
h_1 = v_0 t_1 + \frac{1}{2} a_1 t_1^2 = 0 + \frac{1}{2} (19.6 \, \text{m/s}^2)(50^2 \, \text{s}^2) = (9.8)(2500) \, \text{m} = 24,500 \, \text{m}
\]

(\( h_2 \))

- Initial velocity for coasting ascent: \( v_i = v_{\text{burnout}} = 980 \, \text{m/s} \)
- Final velocity at maximum height: \( v_f = 0 \, \text{m/s} \)
- Acceleration: \( a_2 = -9.8 \, \text{m/s}^2 \)

Using the kinematic equation:

\[
v_f^2 = v_i^2 + 2 a_2 \Delta y
\]

Solving for \( \Delta y \):

\[
0 = (980 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2) \Delta y
\]

\[
0 = 960,400 \, \text{m}^2/\text{s}^2 - 19.6 \, \text{m/s}^2 \times \Delta y
\]

\[
19.6 \, \text{m/s}^2 \times \Delta y = 960,400 \, \text{m}^2/\text{s}^2
\]

\[
\Delta y = \frac{960,400}{19.6} \, \text{m} = 49,000 \, \text{m}
\]

(\( h_{\text{max}} \))

\[
h_{\text{max}} = h_1 + h_2 = 24,500 \, \text{m} + 49,000 \, \text{m} = \mathbf{73,500 \, \text{m}}
\]

(\( t_1 \))

\[
t_1 = 50 \, \text{s}
\]

(\( t_2 \))

Using the kinematic equation:

\[
v_f = v_i + a_2 t_2
\]

Solving for \( t_2 \):

\[
0 = 980 \, \text{m/s} + (-9.8 \, \text{m/s}^2) t_2
\]

\[
9.8 \, \text{m/s}^2 \times t_2 = 980 \, \text{m/s}
\]

\[
t_2 = \frac{980 \, \text{m/s}}{9.8 \, \text{m/s}^2} = 100 \, \text{s}
\]

(\( t_3 \))

Using the kinematic equation:

\[
h_{\text{max}} = \frac{1}{2} g t_3^2
\]

Solving for \( t_3 \):

\[
73,500 \, \text{m} = \frac{1}{2} (9.8 \, \text{m/s}^2) t_3^2
\]

\[
t_3^2 = \frac{73,500 \times 2}{9.8} \, \text{s}^2 = \frac{147,000}{9.8} \, \text{s}^2 = 15,000 \, \text{s}^2
\]

\[
t_3 = \sqrt{15,000} \, \text{s} \approx 122.47 \, \text{s}
\]

(\( t_{\text{total}} \))

\[
t_{\text{total}} = t_1 + t_2 + t_3 = 50 \, \text{s} + 100 \, \text{s} + 122.47 \, \text{s} \approx \mathbf{272.47 \, \text{s}}
\]

- From \( t = 0 \) to \( t = 50 \, \text{s} \), velocity increases linearly from \( 0 \) to \( 980 \, \text{m/s} \).
- From \( t = 50 \, \text{s} \) to \( t = 150 \, \text{s} \), velocity decreases linearly from \( 980 \, \text{m/s} \) to \( 0 \, \text{m/s} \).
- From \( t = 150 \, \text{s} \) to \( t \approx 272.47 \, \text{s} \), velocity decreases linearly from \( 0 \, \text{m/s} \) to \( -1,200 \, \text{m/s} \) (approximate impact velocity).The calculations assume \( g = 9.8 \, \text{m/s}^2 \) and neglect air resistance and changes in \( g \) with altitude, as per the problem statement.

Problem 3/167:

Two race cars have constant accelerations a₁ and a₂ and maximum speeds v₁ and v₂ respectively. They start from rest together and both attain maximum speeds before the finish. The second car reaches the finish line at a time t before the first. What is the length of the track? (a₁ > a₂, v₂ > v₁).

Piecing together the puzzle

I’m working through the length of the track using the cars' accelerations, maximum speeds, and the time difference between their finishes.

Analyzing car dynamics

I’m detailing each car's motion phases: acceleration and constant speed, given their maximum speeds. This helps me understand their time to finish.

Analyzing distance

I’m looking at the total distance traveled by Car 1 and Car 2. Each car has an acceleration phase and a constant speed phase, with Car 2 arriving at the finish line at a time t before Car 1.

Expressing quantities

I’m working through T₁ and T₂, carefully substituting known values to simplify the expressions. This helps in making calculations more manageable for further steps.

Equating times

I’m equating the travel times of Car 1 and Car 2, considering their initial positions, velocities, acceleration, and a time interval t. This involves expressing travel times T₁​ and T₂​ with these parameters.

Calculating time differences

I’m working through the time difference T₁ - T₂ which involves initial velocities, accelerations, and distance. Simplifying the expression, I express it as Δtₐ and Δt𝑐, and I’m now focusing on Δt𝑐, which involves velocities and distances.

Breaking down the equation

I'm simplifying the expression by combining terms involving \( v_1 / a_1 \) and \( v_1 / (2a_1) \), leading to \( v_1 / (2a_1) \).

Weeding out elements

I'm examining the terms in the expression to ensure that \( v_1 / (2 a_1) - v_2 / (2 a_2) \) cancels out, simplifying the overall expression to \( L \left( \frac{1}{v_1} - \frac{1}{v_2} \right) \).

I’m examining the terms in the expression to ensure that v₁ / (2 a₁) - v₂ / (2 a₂) cancels out, simplifying the overall expression to L (1 / v₁ - 1 / v₂).

Breaking down the expression

I’m examining the total difference expression by splitting it into segments, simplifying each, and identifying terms.

Clarifying terms

Piecing together the elements, simplifying, and noting the doubling of terms.

Streamlining the equation

I’m working through simplifying the equation by combining terms and canceling out common factors to make the expression clearer and more manageable. ध्यान देना और स्पष्टता बनाना। ("Pay attention and create clarity" -ed.)

Breaking down the time equations for Car 1 and Car 2, I focus on simplifying terms involving velocities, accelerations, and distance. The final expression for the time difference t in terms of L is derived, incorporating ratios of velocities and accelerations.

Revisiting the equation

It's interesting to see that by rearranging and simplifying the equation, the track length L becomes much clearer:

$$L = [t + (\frac{v_2}{a_2} - \frac{v_1}{a_1})] \times (\frac{v_1}{v_2} - v_2) \quad \text{where:} \quad v_2 > v_1 \quad \text{and} \quad a_1 > a_2$$

Taking a closer look

I'm working through an expression—it had a minus sign in earlier stages. This makes me think it’s crucial for the calculation, so I’m rechecking its importance.

Revisiting the equation

Taking a closer look at the equation for the track length, I’m adjusting the brackets to ensure a positive outcome for L, aiming for better clarity and accuracy.

Crunching numbers

It's intriguing to see that (v₁ / (2 a₁) - v₂ / (2 a₂)) could go either way, but ultimately it depends on the exact values of a₁, a₂, v₁, and v₂. Moving to a different method might clarify things.

Rearranging and rewriting

Piecing together velocity terms and factoring out L, I aim for a cleaner solution. The final expression emerged with careful algebraic manipulation.

Me: What is the answer?

To solve this problem, we'll analyze the motion of both cars separately and then relate them using the given information. Here's a step-by-step solution:

---

### **Understanding the Problem:**

- **Given:**
- Two race cars start from rest and accelerate uniformly.
- Car 1 has acceleration \( a_1 \) and maximum speed \( v_1 \).
- Car 2 has acceleration \( a_2 \) and maximum speed \( v_2 \).
- Both cars reach their maximum speeds before finishing the race.
- Car 2 finishes the race \( t \) seconds before Car 1.
- It's given that \( a_1 > a_2 \) and \( v_2 > v_1 \).

- **Objective:**
- Find the length \( S \) of the track in terms of \( a_1, a_2, v_1, v_2, \) and \( t \).

---

### **Step 1: Analyze Each Car's Motion**

Each car has two phases:

1. **Acceleration Phase:**
- From rest to maximum speed.
- Time to reach maximum speed (\( t_1 \) for Car 1, \( t_2 \) for Car 2).
- Distance covered during acceleration (\( s_{1a} \) and \( s_{2a} \)).

2. **Constant Speed Phase:**
- From attaining maximum speed to finishing the race.
- Time at constant speed (\( t_{1s} \) and \( t_{2s} \)).
- Distance covered during constant speed (\( s_{1s} \) and \( s_{2s} \)).

---

### **Step 2: Calculate Acceleration Phase Parameters**

**For Car 1:**

1. **Time to Reach Maximum Speed (\( t_1 \)):**
\[
v_1 = a_1 t_1 ==> t_1 = \frac{v_1}{a_1}
\]

2. **Distance Covered During Acceleration (\( s_{1a} \)):**
\[
s_{1a} = \frac{1}{2} a_1 t_1^2 = \frac{1}{2} a_1 \left( \frac{v_1}{a_1} \right)^2 = \frac{v_1^2}{2a_1}
\]

**For Car 2:**

1. **Time to Reach Maximum Speed (\( t_2 \)):**
\[
v_2 = a_2 t_2 ==> t_2 = \frac{v_2}{a_2}
\]

2. **Distance Covered During Acceleration (\( s_{2a} \)):**
\[
s_{2a} = \frac{1}{2} a_2 t_2^2 = \frac{1}{2} a_2 \left( \frac{v_2}{a_2} \right)^2 = \frac{v_2^2}{2a_2}
\]

---

### **Step 3: Calculate Total Time and Distance for Each Car**

**Total Distance for Each Car (\( S \)):**

Since both cars travel the same track length:
\[
S = s_{1a} + s_{1s} = s_{2a} + s_{2s}
\]

**Total Time for Each Car:**

- **Car 1:**
\[
T_1 = t_1 + t_{1s}
\]

- **Car 2:**
\[
T_2 = t_2 + t_{2s}
\]

Given that Car 2 finishes \( t \) seconds before Car 1:
\[
T_1 = T_2 + t
\]

---

### **Step 4: Express Constant Speed Phases**

**Distance at Constant Speed:**

- **Car 1:**
\[
s_{1s} = v_1 t_{1s}
\]

- **Car 2:**
\[
s_{2s} = v_2 t_{2s}
\]

**Express \( t_{1s} \) and \( t_{2s} \) in Terms of \( S \):**

From \( S = s_{1a} + s_{1s} \):
\[
t_{1s} = \frac{S - s_{1a}}{v_1} = \frac{S - \frac{v_1^2}{2a_1}}{v_1}
\]

Similarly for Car 2:
\[
t_{2s} = \frac{S - \frac{v_2^2}{2a_2}}{v_2}
\]

---

### **Step 5: Relate the Times of Both Cars**

From the total times:
\[
T_1 = T_2 + t
\]

Substitute \( T_1 \) and \( T_2 \):
\[
t_1 + t_{1s} = t_2 + t_{2s} + t
\]

Substitute \( t_1 \) and \( t_2 \):
\[
\frac{v_1}{a_1} + t_{1s} = \frac{v_2}{a_2} + t_{2s} + t
\]

---

### **Step 6: Substitute \( t_{1s} \) and \( t_{2s} \)**

Replace \( t_{1s} \) and \( t_{2s} \) with expressions from Step 4:
\[
\frac{v_1}{a_1} + \frac{S - \frac{v_1^2}{2a_1}}{v_1} = \frac{v_2}{a_2} + \frac{S - \frac{v_2^2}{2a_2}}{v_2} + t
\]

---

### **Step 7: Simplify the Equation**

1. **Left Side Simplification:**
\[
\frac{v_1}{a_1} + \frac{S}{v_1} - \frac{v_1}{2a_1} = \frac{v_1}{a_1} - \frac{v_1}{2a_1} + \frac{S}{v_1} = \frac{v_1}{2a_1} + \frac{S}{v_1}
\]

2. **Right Side Simplification:**
\[
\frac{v_2}{a_2} + \frac{S}{v_2} - \frac{v_2}{2a_2} + t = \frac{v_2}{2a_2} + \frac{S}{v_2} + t
\]

3. **Set Simplified Sides Equal:**
\[
\frac{v_1}{2a_1} + \frac{S}{v_1} = \frac{v_2}{2a_2} + \frac{S}{v_2} + t
\]

---

### **Step 8: Solve for \( S \)**

1. **Bring \( \frac{S}{v_2} \) to Left Side:**
\[
\frac{S}{v_1} - \frac{S}{v_2} = \frac{v_2}{2a_2} - \frac{v_1}{2a_1} + t
\]

2. **Factor Out \( S \):**
\[
S \left( \frac{1}{v_1} - \frac{1}{v_2} \right) = \frac{v_2}{2a_2} - \frac{v_1}{2a_1} + t
\]

3. **Compute the Difference of Reciprocals:**
\[
\frac{1}{v_1} - \frac{1}{v_2} = \frac{v_2 - v_1}{v_1 v_2}
\]

4. **Solve for \( S \):**
\[
S = \left( \frac{v_1 v_2}{v_2 - v_1} \right) \left( \frac{v_2}{2a_2} - \frac{v_1}{2a_1} + t \right)
\]

---

### **Final Answer:**

**The length \( S \) of the track is:**

\[
{S = \left( \frac{v_1 v_2}{v_2 - v_1} \right) \left( \frac{v_2}{2a_2} - \frac{v_1}{2a_1} + t \right)}
\]

---

### **Explanation:**

- **\( \frac{v_1 v_2}{v_2 - v_1} \):** This term accounts for the relative speeds of the two cars.
- **\( \frac{v_2}{2a_2} - \frac{v_1}{2a_1} + t \):** This term represents the difference in times due to acceleration and the given time difference \( t \).

**Note:** This formula provides the track length \( S \) in terms of the given accelerations \( a_1, a_2 \), maximum speeds \( v_1, v_2 \), and time difference \( t \).

Problem 4/167:

A column of troops 3 miles long marches steadily down a straight road. An officer rides at a constant rate from the rear of the column to the head and immediately back to the rear. If the officer travels at a speed equal to twice that of the troops, how much of an advance will the troops make during one such circuit by the officer?

To solve this problem, we'll determine the time it takes for the officer to complete his circuit (from the rear to the head and back to the rear) and then calculate how much distance the troops cover during this time.

**Definitions:**

- \( L = 3 \) miles (length of the column)
- \( v \) = speed of the troops
- Officer's speed \( = 2v \)

**Step 1: Time from Rear to Head (First Leg)**

The officer and the head of the column are both moving forward. The officer's relative speed towards the head is:

\[
\text{Relative speed} = \text{Officer's speed} - \text{Troops' speed} = 2v - v = v
\]

Time taken to reach the head:

\[
t_1 = \frac{\text{Distance to head}}{\text{Relative speed}} = \frac{L}{v}
\]

**Step 2: Time from Head Back to Rear (Second Leg)**

When the officer returns, he moves backward relative to the ground at speed \( 2v \), while the rear of the column moves forward at speed \( v \). The relative speed between the officer and the rear is:

\[
\text{Relative speed} = 2v + v = 3v
\]

Time taken to return to the rear:

\[
t_2 = \frac{\text{Distance between head and rear at time } t_1}{\text{Relative speed}}
\]

At time \( t_1 \), the distance between the head and rear is:

\[
\text{Distance} = x_{\text{head}}(t_1) - x_{\text{rear}}(t_1) = (L + vt_1) - (vt_1) = L
\]

So,

\[
t_2 = \frac{L}{3v}
\]

**Step 3: Total Time and Distance Advanced by Troops**

Total time \( T = t_1 + t_2 \):

\[
T = \frac{L}{v} + \frac{L}{3v} = \frac{4L}{3v}
\]

Distance advanced by the troops:

\[
\text{Distance} = v \times T = v \times \frac{4L}{3v} = \frac{4L}{3}
\]

Substituting \( L = 3 \) miles:

\[
\text{Distance} = \frac{4 \times 3}{3} = 4 \text{ miles}
\]

**Answer:**

4 miles

Problem 5/167:

On the long horizontal test track at Edwards AFB, both rocket and jet motors can be tested. On a certain day, a rocket motor, started from rest, accelerated constantly until its fuel was exhausted, after which it ran at constant speed. It was observed that this exhaustion of the rocket fuel took place as the rocket passed the mid-point of the measured test distance. Then a jet motor was started from rest down the track, with a constant acceleration for the entire distance. It was observed that both rocket and jet motors covered the test distance in exactly the same time. What was the ratio of the acceleration of the jet motor to that of the rocket motor?

To solve this problem, we'll analyze the motion of both the rocket and the jet using kinematic equations. Let's define the total test distance as \( D \). The rocket and jet both cover this distance in the same total time \( T \).

**Rocket Motor:**

1. **First Half (\( 0 \) to \( D/2 \))**:
- Initial velocity (\( u \)) = 0
- Acceleration (\( a_r \)) = acceleration of the rocket
- Distance (\( s \)) = \( D/2 \)

Using the kinematic equation:
\[
v_1^2 = u^2 + 2a_r s ==> v_1^2 = 0 + 2a_r \left( \frac{D}{2} \right) ==> v_1 = \sqrt{a_r D}
\]
Where \( v_1 \) is the velocity at \( D/2 \).

2. **Second Half (\( D/2 \) to \( D \))**:
- The rocket moves at constant speed \( v_1 \).
- Time to cover the second half:
\[
t_2 = \frac{D/2}{v_1} = \frac{D}{2\sqrt{a_r D}} = \frac{\sqrt{D}}{2\sqrt{a_r}}
\]

3. **Time for First Half**:
\[
v_1 = a_r t_1 ==> t_1 = \frac{v_1}{a_r} = \frac{\sqrt{a_r D}}{a_r} = \frac{\sqrt{D}}{\sqrt{a_r}}
\]

4. **Total Time for Rocket**:
\[
T_r = t_1 + t_2 = \frac{\sqrt{D}}{\sqrt{a_r}} + \frac{\sqrt{D}}{2\sqrt{a_r}} = \frac{3\sqrt{D}}{2\sqrt{a_r}}
\]

**Jet Motor:**

- Accelerates uniformly from rest over the distance \( D \):
\[
D = \frac{1}{2} a_j T_j^2
\]
- Since \( T_j = T_r \):
\[
D = \frac{1}{2} a_j \left( \frac{3\sqrt{D}}{2\sqrt{a_r}} \right)^2 = \frac{1}{2} a_j \left( \frac{9D}{4a_r} \right)
\]
- Simplify:
\[
D = \frac{9D a_j}{8a_r}
\]
- Solve for the ratio \( \frac{a_j}{a_r} \):
\[
1 = \frac{9 a_j}{8 a_r} ==> \frac{a_j}{a_r} = \frac{8}{9}
\]

**Answer:**

\({\frac{8}{9}}\)