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Sagot :
Certainly! Let's break down the problem step-by-step and find the instantaneous velocity, average velocity, and distance fallen for a freely falling object 10 seconds after it is released from rest. We will ignore air resistance for this calculation.
### Given Data:
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- Time, [tex]\( t = 10 \, \text{seconds} \)[/tex]
- Initial velocity, [tex]\( u = 0 \, \text{m/s} \)[/tex] (since it is released from rest)
### Instantaneous Velocity after 10 seconds
The instantaneous velocity [tex]\( v \)[/tex] of a freely falling object can be calculated using the equation of motion:
[tex]\[ v = u + g \cdot t \][/tex]
Since the initial velocity [tex]\( u \)[/tex] is 0, the equation simplifies to:
[tex]\[ v = g \cdot t \][/tex]
Plugging in the given values:
[tex]\[ v = 9.8 \, \text{m/s}^2 \times 10 \, \text{seconds} \][/tex]
[tex]\[ v = 98 \, \text{m/s} \][/tex]
So, the instantaneous velocity after 10 seconds is [tex]\( 98 \, \text{m/s} \)[/tex].
### Average Velocity during the 10-second interval
The average velocity [tex]\( v_{\text{avg}} \)[/tex] of a freely falling object can be computed as the average of the initial and final velocities:
[tex]\[ v_{\text{avg}} = \frac{u + v}{2} \][/tex]
Since [tex]\( u = 0 \, \text{m/s} \)[/tex] and [tex]\( v = 98 \, \text{m/s} \)[/tex]:
[tex]\[ v_{\text{avg}} = \frac{0 + 98}{2} \][/tex]
[tex]\[ v_{\text{avg}} = \frac{98}{2} \][/tex]
[tex]\[ v_{\text{avg}} = 49 \, \text{m/s} \][/tex]
So, the average velocity during the 10-second interval is [tex]\( 49 \, \text{m/s} \)[/tex].
### Distance Fallen during this time
The distance fallen [tex]\( s \)[/tex] by the object can be calculated using the equation of motion:
[tex]\[ s = ut + \frac{1}{2} g t^2 \][/tex]
Since the initial velocity [tex]\( u \)[/tex] is 0, the equation simplifies to:
[tex]\[ s = \frac{1}{2} g t^2 \][/tex]
Plugging in the given values:
[tex]\[ s = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (10 \, \text{seconds})^2 \][/tex]
[tex]\[ s = \frac{1}{2} \times 9.8 \times 100 \][/tex]
[tex]\[ s = 4.9 \times 100 \][/tex]
[tex]\[ s = 490 \, \text{meters} \][/tex]
So, the distance fallen during this 10-second interval is [tex]\( 490 \, \text{meters} \)[/tex].
### Summary:
1. The instantaneous velocity after 10 seconds is [tex]\( 98 \, \text{m/s} \)[/tex].
2. The average velocity during the 10-second interval is [tex]\( 49 \, \text{m/s} \)[/tex].
3. The distance fallen during this time is [tex]\( 490 \, \text{meters} \)[/tex].
### Given Data:
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- Time, [tex]\( t = 10 \, \text{seconds} \)[/tex]
- Initial velocity, [tex]\( u = 0 \, \text{m/s} \)[/tex] (since it is released from rest)
### Instantaneous Velocity after 10 seconds
The instantaneous velocity [tex]\( v \)[/tex] of a freely falling object can be calculated using the equation of motion:
[tex]\[ v = u + g \cdot t \][/tex]
Since the initial velocity [tex]\( u \)[/tex] is 0, the equation simplifies to:
[tex]\[ v = g \cdot t \][/tex]
Plugging in the given values:
[tex]\[ v = 9.8 \, \text{m/s}^2 \times 10 \, \text{seconds} \][/tex]
[tex]\[ v = 98 \, \text{m/s} \][/tex]
So, the instantaneous velocity after 10 seconds is [tex]\( 98 \, \text{m/s} \)[/tex].
### Average Velocity during the 10-second interval
The average velocity [tex]\( v_{\text{avg}} \)[/tex] of a freely falling object can be computed as the average of the initial and final velocities:
[tex]\[ v_{\text{avg}} = \frac{u + v}{2} \][/tex]
Since [tex]\( u = 0 \, \text{m/s} \)[/tex] and [tex]\( v = 98 \, \text{m/s} \)[/tex]:
[tex]\[ v_{\text{avg}} = \frac{0 + 98}{2} \][/tex]
[tex]\[ v_{\text{avg}} = \frac{98}{2} \][/tex]
[tex]\[ v_{\text{avg}} = 49 \, \text{m/s} \][/tex]
So, the average velocity during the 10-second interval is [tex]\( 49 \, \text{m/s} \)[/tex].
### Distance Fallen during this time
The distance fallen [tex]\( s \)[/tex] by the object can be calculated using the equation of motion:
[tex]\[ s = ut + \frac{1}{2} g t^2 \][/tex]
Since the initial velocity [tex]\( u \)[/tex] is 0, the equation simplifies to:
[tex]\[ s = \frac{1}{2} g t^2 \][/tex]
Plugging in the given values:
[tex]\[ s = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (10 \, \text{seconds})^2 \][/tex]
[tex]\[ s = \frac{1}{2} \times 9.8 \times 100 \][/tex]
[tex]\[ s = 4.9 \times 100 \][/tex]
[tex]\[ s = 490 \, \text{meters} \][/tex]
So, the distance fallen during this 10-second interval is [tex]\( 490 \, \text{meters} \)[/tex].
### Summary:
1. The instantaneous velocity after 10 seconds is [tex]\( 98 \, \text{m/s} \)[/tex].
2. The average velocity during the 10-second interval is [tex]\( 49 \, \text{m/s} \)[/tex].
3. The distance fallen during this time is [tex]\( 490 \, \text{meters} \)[/tex].
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