A runner accelerates at a constant 0.20 m/s2 to the north.
If she starts from rest, how long does it take her to reach 7.0 m/s? How far does she run in this time?
If she was initially running at 3.0 m/s to the north, how long does it take her to run 50 meters? What is her speed at that time?
If she was initially running at 3.0 m/s to the north, how fast is she moving after running 30 meters?
If she was initially moving at 4.0 m/s to the South, how long does it take her to return to her starting position?
If she was initially moving at 4.0 m/s to the South, how long does it take her to reach a point 60 meters South of her starting position?
The correct answer and explanation is :
To analyze the runner’s motion under constant acceleration, we employ the kinematic equations:
- ( v = u + at )
- ( s = ut + \frac{1}{2}at^2 )
- ( v^2 = u^2 + 2as )
Where:
- ( u ) is the initial velocity
- ( v ) is the final velocity
- ( a ) is the acceleration
- ( t ) is the time
- ( s ) is the displacement
1. Starting from rest, reaching 7.0 m/s:
- Time to reach 7.0 m/s: Using ( v = u + at ): ( 7.0\, \text{m/s} = 0 + (0.20\, \text{m/s}^2) \times t ) Solving for ( t ): ( t = \frac{7.0\, \text{m/s}}{0.20\, \text{m/s}^2} = 35\, \text{seconds} )
- Distance covered in this time: Using ( s = ut + \frac{1}{2}at^2 ): ( s = 0 + \frac{1}{2}(0.20\, \text{m/s}^2)(35\, \text{s})^2 ) ( s = 0.10 \times 1,225 = 122.5\, \text{meters} )
2. Initially running at 3.0 m/s, covering 50 meters:
- Time to cover 50 meters: Using ( s = ut + \frac{1}{2}at^2 ): ( 50\, \text{m} = (3.0\, \text{m/s})t + \frac{1}{2}(0.20\, \text{m/s}^2)t^2 ) This simplifies to a quadratic equation: ( 0.1t^2 + 3.0t – 50 = 0 ) Solving for ( t ) using the quadratic formula: ( t = \frac{-3.0 \pm \sqrt{(3.0)^2 – 4 \times 0.1 \times (-50)}}{2 \times 0.1} ) ( t = \frac{-3.0 \pm \sqrt{9 + 20}}{0.2} ) ( t = \frac{-3.0 \pm \sqrt{29}}{0.2} )Taking the positive root:( t = \frac{-3.0 + 5.385}{0.2} \approx 11.925\, \text{seconds} )
- Speed at that time: Using ( v = u + at ): ( v = 3.0\, \text{m/s} + (0.20\, \text{m/s}^2) \times 11.925\, \text{s} ) ( v \approx 3.0 + 2.385 = 5.385\, \text{m/s} )
3. Initially running at 3.0 m/s, covering 30 meters:
- Final speed after 30 meters: Using ( v^2 = u^2 + 2as ): ( v^2 = (3.0\, \text{m/s})^2 + 2(0.20\, \text{m/s}^2)(30\, \text{m}) ) ( v^2 = 9 + 12 ) ( v = \sqrt{21} \approx 4.582\, \text{m/s} )
4. Initially moving at 4.0 m/s to the South, returning to starting position:
Since the runner is moving southward initially, to return to the starting position, she must decelerate to a stop and then accelerate northward back to the start.
- Time to stop: Using ( v = u + at ) with ( v = 0 ): ( 0 = 4.0\, \text{m/s} – (0.20\, \text{m/s}^2)t ) ( t = \frac{4.0\, \text{m/s}}{0.20\, \text{m/s}^2} = 20\, \text{seconds} )
- Distance covered while stopping: Using ( s = ut + \frac{1}{2}at^2 ): ( s = (4.0\, \text{m/s})(20\, \text{s}) – \frac{1}{2}(0.20\, \text{m/s}^2)(20\, \text{s})^2 ) ( s = 80 – 40 = 40\, \text{meters} ) The runner has moved 40 meters south while stopping. To return to the starting position, she needs to cover 40 meters north.
- Time to accelerate northward back to the start: Using ( s = ut + \frac{1}{2}at^2 ) with