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).