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:
- Final velocity (( v )): [ v = v_0 + at ]
- Displacement (( s )): [ s = v_0 t + \frac{1}{2} a t^2 ]
- Velocity-displacement relation: [ v^2 = v_0^2 + 2a s ]
where:
- ( v_0 ) is the initial velocity
- ( a ) is the constant acceleration
- ( t ) is the time elapsed
- ( s ) is the displacement
1. Reaching 7.0 m/s from rest with ( a = 0.20 \, \text{m/s}^2 ):
- Time to reach 7.0 m/s: Using ( v = v_0 + at ): [ t = \frac{v – v_0}{a} = \frac{7.0 \, \text{m/s} – 0}{0.20 \, \text{m/s}^2} = 35 \, \text{s} ]
- Distance covered in this time: Using ( s = v_0 t + \frac{1}{2} a t^2 ): [ s = 0 \times 35 \, \text{s} + \frac{1}{2} \times 0.20 \, \text{m/s}^2 \times (35 \, \text{s})^2 = 428.75 \, \text{m} ]
2. Covering 50 meters starting at 3.0 m/s:
- Time to cover 50 meters: Using ( s = v_0 t + \frac{1}{2} a t^2 ): [ 50 \, \text{m} = 3.0 \, \text{m/s} \times t + \frac{1}{2} \times 0.20 \, \text{m/s}^2 \times t^2 ] This simplifies to: [ 0.1 t^2 + 3.0 t – 50 = 0 ] Solving this quadratic equation: [ 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} = \frac{-3.0 \pm \sqrt{29}}{0.2} ] Taking the positive root: [ t \approx \frac{-3.0 + 5.385}{0.2} = \frac{2.385}{0.2} \approx 11.925 \, \text{s} ]
- Speed at that time: Using ( v = v_0 + at ): [ v = 3.0 \, \text{m/s} + 0.20 \, \text{m/s}^2 \times 11.925 \, \text{s} \approx 5.385 \, \text{m/s} ]
3. Speed after running 30 meters starting at 3.0 m/s:
Using ( v^2 = v_0^2 + 2a s ):
[ 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. Returning to the starting position after moving south at 4.0 m/s:
Assuming southward direction as negative:
Initial velocity, ( v_0 = -4.0 \, \text{m/s} )
The runner decelerates until velocity becomes zero, then accelerates northward back to the starting point.
- Time to stop: Using ( v = v_0 + at ):[ 0 = -4.0 \, \text{m/s} + 0.20 \, \text{m/s}^2 \times t ] [ t = \frac{4.0 \, \text{m/s}}{0.20 \, \text{m/s}^2} = 20 \, \text{s} ]
- Distance covered while stopping: Using ( s = v_0 t + \frac{1}{2} a t^2 ): [ s = -4.0 \, \text{m/s} \times 20 \, \text{s} + \frac{1}{2} \times 0.20 \, \text{m/s}^2 \times (20 \, \text{s})^2 ] [ s = -80 \, \text{m} + 40 \, \text{m} = -40 \, \text{m} ]