Question

1. A car, initially traveling east with a speed of 5.0 meters per second, is accelerated uniformly at 2.0 meters per second 2 east for 10 seconds along a straight line. What is the total distance traveled by the car during this 10-second interval? 2. If a car accelerates uniformly from rest to 15 meters per second over a distance of 100 meters, the magnitude of the car's acceleration is 1. 0.15 m/s 2 2. 2.3 m/s 2 3. 1.1 m/s 2 4. 6.7 m/s 2 © Success Academy Charter Schools1
3. A car traveling on a straight road at 15.0 meters per second accelerates uniformly to a speed of 21.0 meters per second in 12.0 seconds. What is the total distance traveled by the car in this 12.0-second time interval? 4. A 747 jet traveling at a velocity of 70 meters per second north touches down on a runway. The jet slows to rest at the rate of 2.0 meters per second 2 . Calculate the total distance the jet travels on the runway as it is brought to rest. 5. Landing with a speed of 71.4 m/s, and traveling due south, a jet comes to rest in 949 m. Assuming the jet slows with constant acceleration, find the magnitude and direction of its acceleration.

Answer 1

Let's solve each problem step by step.

### Problem 1:
**Given:**
- Initial speed, \( v_0 = 5.0 \, \text{m/s} \) (east)
- Acceleration, \( a = 2.0 \, \text{m/s}^2 \) (east)
- Time, \( t = 10 \, \text{s} \)

**Distance traveled can be found using the equation:**
\[
d = v_0 t + \frac{1}{2} a t^2
\]

**Substituting the values:**
\[
d = (5.0 \, \text{m/s}) \cdot (10 \, \text{s}) + \frac{1}{2} (2.0 \, \text{m/s}^2) \cdot (10 \, \text{s})^2
\]
\[
d = 50.0 \, \text{m} + \frac{1}{2} \cdot 2.0 \cdot 100
\]
\[
d = 50.0 \, \text{m} + 100.0 \, \text{m}
\]
\[
d = 150.0 \, \text{m}
\]

**Answer: The total distance traveled by the car is 150.0 meters.**

---

### Problem 2:
**Given:**
- Initial velocity, \( v_0 = 0 \, \text{m/s} \) (starts from rest)
- Final velocity, \( v = 15 \, \text{m/s} \)
- Distance, \( d = 100 \, \text{m} \)

**Using the equation of motion:**
\[
v^2 = v_0^2 + 2ad
\]
Substituting the values:
\[
(15 \, \text{m/s})^2 = 0 + 2a(100 \, \text{m})
\]
\[
225 = 200a
\]
\[
a = \frac{225}{200} = 1.125 \, \text{m/s}^2
\]

**Answer: The magnitude of the car's acceleration is approximately 1.1 m/s². (Option 3)**

---

### Problem 3:
**Given:**
- Initial speed, \( v_0 = 15.0 \, \text{m/s} \)
- Final speed, \( v = 21.0 \, \text{m/s} \)
- Time, \( t = 12.0 \, \text{s} \)

**First, we find the acceleration:**
\[
a = \frac{v - v_0}{t} = \frac{21.0 - 15.0}{12.0} = \frac{6.0}{12.0} = 0.5 \, \text{m/s}^2
\]

**Now, calculate the total distance using:**
\[
d = v_0 t + \frac{1}{2} a t^2
\]
\[
d = (15.0 \, \text{m/s}) (12.0 \, \text{s}) + \frac{1}{2} (0.5 \, \text{m/s}^2)(12.0 \, \text{s})^2
\]
\[
d = 180.0 \, \text{m} + \frac{1}{2} (0.5) (144)
\]
\[
d = 180.0 \, \text{m} + 36.0 \, \text{m}
\]
\[
d = 216.0 \, \text{m}
\]

**Answer: The total distance traveled by the car in this time interval is 216.0 meters.**

---

### Problem 4:
**Given:**
- Initial speed, \( v_0 = 70 \, \text{m/s} \)
- Final speed, \( v = 0 \, \text{m/s} \)
- Deceleration, \( a = -2.0 \, \text{m/s}^2 \) (the negative sign indicates slowing down)

**Using the following equation:**
\[
v^2 = v_0^2 + 2ad
\]
Substituting the values:
\[
0 = (70)^2 + 2(-2.0)d
\]
\[
0 = 4900 - 4d
\]
\[
4d = 4900 \implies d = \frac{4900}{4} = 1225 \, \text{m}
\]

**Answer: The total distance the jet travels on the runway is 1225 meters.**

---

### Problem 5:
**Given:**
- Initial speed, \( v_0 = 71.4 \, \text{m/s} \) (south)
- Final speed, \( v = 0 \, \text{m/s} \)
- Distance, \( d = 949 \, \text{m} \)

**Using the equation of motion again:**
\[
v^2 = v_0^2 + 2ad
\]
\[
0 = (71.4)^2 + 2a(949)
\]
Calculating \( (71.4)^2 \):
\[
0 = 5097.96 + 1898a
\]
\[
1898a = -5097.96
\]
\[
a = \frac{-5097.96}{1898} \approx -2.68 \, \text{m/s}^2
\]

**Answer: The magnitude of the acceleration is approximately 2.68 m/s², directed north (opposite to the motion).**