SOLUTION: An automobile accelerates from rest at 1.7 m/s2 for 15.5 s. The speed is then held constant for 23.4 s, after which there is an acceleration of -3.5 m/s2 until the automobile stops

Question 1165645: An automobile accelerates from rest at 1.7 m/s2 for 15.5 s. The speed is then held constant for 23.4 s, after which there is an acceleration of -3.5 m/s2 until the automobile stops. What is the total distance traveled?
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
# Given parameters
a1 = 1.7 # m/s^2
t1 = 15.5 # s
v0_1 = 0 # m/s
a3 = -3.5 # m/s^2
v_final_3 = 0 # m/s
t2 = 23.4 # s
# --- Phase 1: Acceleration ---
# 1. Final velocity of Phase 1 (v_f1), which is the constant velocity for Phase 2 (v2)
v_f1 = v0_1 + a1 * t1
v2 = v_f1
# 2. Distance traveled in Phase 1 (d1)
d1 = v0_1 * t1 + 0.5 * a1 * t1**2
# --- Phase 2: Constant Speed ---
# 3. Distance traveled in Phase 2 (d2)
d2 = v2 * t2
# --- Phase 3: Deceleration ---
# 4. Initial velocity of Phase 3 (v_03) is v_f1
v0_3 = v_f1
# Distance traveled in Phase 3 (d3): v_f^2 = v_0^2 + 2*a*d
# 0 = v_f1^2 + 2*a3*d3
d3 = -(v_f1**2) / (2 * a3)
# --- Total Distance ---
# 5. Total distance (D_total)
D_total = d1 + d2 + d3
print(f"Final velocity of Phase 1 (v_f1): {v_f1} m/s")
print(f"Distance in Phase 1 (d1): {d1} m")
print(f"Distance in Phase 2 (d2): {d2} m")
print(f"Distance in Phase 3 (d3): {d3} m")
print(f"Total Distance (D_total): {D_total} m")
```
```text?code_stdout&code_event_index=2
Final velocity of Phase 1 (v_f1): 26.349999999999998 m/s
Distance in Phase 1 (d1): 204.2125 m
Distance in Phase 2 (d2): 616.5899999999999 m
Distance in Phase 3 (d3): 99.18892857142855 m
Total Distance (D_total): 919.9914285714284 m
```
The total distance traveled by the automobile is approximately **$920.0$ meters**.
The movement is calculated in three phases: acceleration, constant speed, and deceleration.
-----
## 1\. Phase 1: Acceleration
The automobile accelerates from rest ($v_0 = 0 \text{ m/s}$) for $15.5 \text{ s}$.
### A. Final Speed ($v_1$)
This speed will be the constant speed for Phase 2.
$$v_1 = v_0 + a_1 t_1$$
$$v_1 = 0 + (1.7 \text{ m/s}^2)(15.5 \text{ s})$$
$$\mathbf{v_1 = 26.35 \text{ m/s}}$$
### B. Distance Traveled ($d_1$)
$$d_1 = v_0 t_1 + \frac{1}{2} a_1 t_1^2$$
$$d_1 = 0 + \frac{1}{2} (1.7 \text{ m/s}^2)(15.5 \text{ s})^2$$
$$\mathbf{d_1 \approx 204.21 \text{ m}}$$
-----
## 2\. Phase 2: Constant Speed
The speed is held constant ($v_2 = 26.35 \text{ m/s}$) for $23.4 \text{ s}$.
### A. Distance Traveled ($d_2$)
$$d_2 = v_2 t_2$$
$$d_2 = (26.35 \text{ m/s})(23.4 \text{ s})$$
$$\mathbf{d_2 \approx 616.59 \text{ m}}$$
-----
## 3\. Phase 3: Deceleration
The automobile decelerates from $v_0 = 26.35 \text{ m/s}$ at $a_3 = -3.5 \text{ m/s}^2$ until it stops ($v_f = 0 \text{ m/s}$).
### A. Distance Traveled ($d_3$)
We use the time-independent kinematic equation:
$$v_f^2 = v_0^2 + 2 a_3 d_3$$
$$0^2 = (26.35 \text{ m/s})^2 + 2 (-3.5 \text{ m/s}^2) d_3$$
$$0 = 694.3225 - 7.0 d_3$$
$$7.0 d_3 = 694.3225$$
$$\mathbf{d_3 \approx 99.19 \text{ m}}$$
-----
## 4\. Total Distance Traveled
The total distance is the sum of the distances from all three phases:
$$D_{total} = d_1 + d_2 + d_3$$
$$D_{total} = 204.21 \text{ m} + 616.59 \text{ m} + 99.19 \text{ m}$$
$$\mathbf{D_{total} \approx 919.99 \text{ m}}$$
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