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 utilize 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:
- Initial velocity (( u )) = 0 m/s
- Final velocity (( v )) = 7.0 m/s
- Acceleration (( a )) = 0.20 m/s²
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 )
( 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} \times 0.20\, \text{m/s}^2 \times (35\, \text{seconds})^2 )
( s = 0.10 \times 1225 = 122.5\, \text{meters} )
2. Initially running at 3.0 m/s, covering 50 meters:
- Initial velocity (( u )) = 3.0 m/s
- Displacement (( s )) = 50 meters
- Acceleration (( a )) = 0.20 m/s²
Time to cover 50 meters:
Using ( s = ut + \frac{1}{2}at^2 ):
( 50\, \text{m} = (3.0\, \text{m/s}) \times t + \frac{1}{2} \times 0.20\, \text{m/s}^2 \times t^2 )
( 50 = 3t + 0.1t^2 )
Rearranging:
( 0.1t^2 + 3t – 50 = 0 )
Solving this quadratic equation:
( t = \frac{-3 \pm \sqrt{(3)^2 – 4 \times 0.1 \times (-50)}}{2 \times 0.1} )
( t = \frac{-3 \pm \sqrt{9 + 20}}{0.2} = \frac{-3 \pm \sqrt{29}}{0.2} )
( t \approx \frac{-3 + 5.385}{0.2} = \frac{2.385}{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{seconds} )
( v \approx 3.0 + 2.385 = 5.385\, \text{m/s} )
3. Initially running at 3.0 m/s, after running 30 meters:
- Initial velocity (( u )) = 3.0 m/s
- Displacement (( s )) = 30 meters
- Acceleration (( a )) = 0.20 m/s²
Final speed after 30 meters:
Using ( v^2 = u^2 + 2as ):
( v^2 = (3.0\, \text{m/s})^2 + 2 \times 0.20\, \text{m/s}^2 \times 30\, \text{m} )
( v^2 = 9 + 12 = 21 )
( v = \sqrt{21} \approx 4.58\, \text{m/s} )
4. Initially moving at 4.0 m/s to the South, returning to starting position:
- Initial velocity (( u )) = -4.0 m/s (negative sign indicates southward direction)
- Acceleration (( a )) = 0.20 m/s² (northward)
Time to return to starting position:
The runner decelerates to a stop and then accelerates northward back to the starting point.
Time to stop:
Using ( v = u + at ):
( 0 = -4.0\, \text{m/s} + 0.20\, \text{m/s}^2 \times t )
( t = \frac{4.0