Problem:
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?
Given:
- 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.
(a) Calculating the Speed at the 0.20 Mile Post:
- Write the equations of motion for $s_1$ and $s_2$: $$s_1 = v_0 t_1 + \tfrac{1}{2} a t_1^2 = 0.10 \text{ miles}$$
$$s_2 = v_0 t_2 + \tfrac{1}{2} a t_2^2 = 0.20 \text{ miles}$$ - Subtract the first equation from the second to eliminate $v_0$: $$s_2 – s_1 = v_0 (t_2 – t_1) + \tfrac{1}{2} a (t_2^2 – t_1^2)$$
$$0.10 = v_0 (8) + \tfrac{1}{2} a (320)$$
$$0.10 = 8v_0 + 160a \quad \text{(Equation 1)}$$ - Express $s_1$ in terms of $v_0$ and $a$: $$0.10 = 16v_0 + 128a \quad \text{(Equation 2)}$$
- Subtract Equation 2 from Equation 1 to solve for $v_0$: $$0 = -8v_0 + 32a$$
$$v_0 = 4a$$ - Substitute $v_0$ back into Equation 2: $$0.10 = 16(4a) + 128a$$
$$0.10 = 192a$$
$$a = \frac{0.10}{192} \text{ miles/sec}^2$$ - Calculate $v_0$: $$v_0 = 4a = \frac{4 \times 0.10}{192} = \frac{0.10}{48} \text{ miles/sec}$$
- Find the velocity at $t_2 = 24$ seconds: $$v = v_0 + a t_2 = \frac{0.10}{48} + \left( \frac{0.10}{192} \times 24 \right)$$
$$v = \frac{0.10}{48} + \frac{0.10 \times 24}{192}$$
$$v = \frac{0.10}{48} + \frac{0.10}{8}$$
$$v = \frac{0.10}{48} + \frac{0.10}{8} = \frac{0.10(1 + 6)}{48} = \frac{0.10 \times 7}{48}$$ - Convert $v$ to miles/hour: $$v = \frac{0.10 \times 7}{48} \times 3600 \text{ miles/hour}$$
$$v = \frac{0.70}{48} \times 3600 \approx 52.5 \text{ miles/hour}$$
Answer for (a): The speedometer should have read 52.5 miles per hour at the 0.20-mile post.
(b) Calculating the Maximum Acceleration:
- Use the calculated acceleration: $$a = \frac{0.10}{192} \text{ miles/sec}^2$$
- Convert miles to feet ($1 \text{ mile} = 5280 \text{ feet}$): $$a = \frac{0.10 \times 5280}{192} \text{ ft/sec}^2$$
$$a = \frac{528}{192} \text{ ft/sec}^2 \approx 2.75 \text{ ft/sec}^2$$
Answer for (b): His maximum acceleration was approximately 2.75 feet per second squared.
Final Answers:
(a) The speedometer should have read 52.5 miles per hour at the 0.20-mile post.
(b) His maximum acceleration was approximately 2.75 ft/sec².