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Sagot :
Let's solve the problem step-by-step.
### (i) Finding the acceleration of the vehicle:
Acceleration is defined as the change in velocity (speed) over time. We are given a list of speeds at different times, and we need to find the overall acceleration.
1. Identify the initial and final speeds:
- Initial speed, [tex]\( v_i \)[/tex]: [tex]\(0 \, \text{m/s}\)[/tex]
- Final speed, [tex]\( v_f \)[/tex]: [tex]\(10 \, \text{m/s}\)[/tex]
2. Identify the initial and final times:
- Initial time, [tex]\( t_i \)[/tex]: [tex]\(0 \, \text{s}\)[/tex]
- Final time, [tex]\( t_f \)[/tex]: [tex]\(5 \, \text{s}\)[/tex]
3. Calculate the change in velocity (Δv):
[tex]\[ \Delta v = v_f - v_i = 10 \, \text{m/s} - 0 \, \text{m/s} = 10 \, \text{m/s} \][/tex]
4. Calculate the change in time (Δt):
[tex]\[ \Delta t = t_f - t_i = 5 \, \text{s} - 0 \, \text{s} = 5 \, \text{s} \][/tex]
5. Use the formula for acceleration (a):
[tex]\[ a = \frac{\Delta v}{\Delta t} = \frac{10 \, \text{m/s}}{5 \, \text{s}} = 2 \, \text{m/s}^2 \][/tex]
So, the acceleration of the vehicle is [tex]\( 2 \, \text{m/s}^2 \)[/tex].
### (ii) Calculating the distance covered in 5 seconds:
To find the distance covered, we can use one of the kinematic equations for uniformly accelerated motion. We know:
- Initial speed, [tex]\( v_i \)[/tex]: [tex]\(0 \, \text{m/s}\)[/tex]
- Acceleration, [tex]\( a \)[/tex]: [tex]\(2 \, \text{m/s}^2\)[/tex]
- Time, [tex]\( t \)[/tex]: [tex]\(5 \, \text{s}\)[/tex]
The kinematic equation that relates these quantities is:
[tex]\[ \text{distance} = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2 \][/tex]
Plug in the known values:
[tex]\[ \text{distance} = 0 \cdot 5 \, \text{s} + \frac{1}{2} \cdot 2 \, \text{m/s}^2 \cdot (5 \, \text{s})^2 \][/tex]
Calculate each term:
[tex]\[ = 0 + \frac{1}{2} \cdot 2 \, \text{m/s}^2 \cdot 25 \, \text{s}^2 \][/tex]
[tex]\[ = \frac{1}{2} \cdot 2 \cdot 25 \][/tex]
[tex]\[ = 25 \, \text{m} \][/tex]
So, the distance covered by the vehicle in 5 seconds is [tex]\( 25 \, \text{meters} \)[/tex].
### Summary:
(i) The acceleration of the vehicle is [tex]\( 2 \, \text{m/s}^2 \)[/tex].
(ii) The distance covered in 5 seconds is [tex]\( 25 \, \text{meters} \)[/tex].
### (i) Finding the acceleration of the vehicle:
Acceleration is defined as the change in velocity (speed) over time. We are given a list of speeds at different times, and we need to find the overall acceleration.
1. Identify the initial and final speeds:
- Initial speed, [tex]\( v_i \)[/tex]: [tex]\(0 \, \text{m/s}\)[/tex]
- Final speed, [tex]\( v_f \)[/tex]: [tex]\(10 \, \text{m/s}\)[/tex]
2. Identify the initial and final times:
- Initial time, [tex]\( t_i \)[/tex]: [tex]\(0 \, \text{s}\)[/tex]
- Final time, [tex]\( t_f \)[/tex]: [tex]\(5 \, \text{s}\)[/tex]
3. Calculate the change in velocity (Δv):
[tex]\[ \Delta v = v_f - v_i = 10 \, \text{m/s} - 0 \, \text{m/s} = 10 \, \text{m/s} \][/tex]
4. Calculate the change in time (Δt):
[tex]\[ \Delta t = t_f - t_i = 5 \, \text{s} - 0 \, \text{s} = 5 \, \text{s} \][/tex]
5. Use the formula for acceleration (a):
[tex]\[ a = \frac{\Delta v}{\Delta t} = \frac{10 \, \text{m/s}}{5 \, \text{s}} = 2 \, \text{m/s}^2 \][/tex]
So, the acceleration of the vehicle is [tex]\( 2 \, \text{m/s}^2 \)[/tex].
### (ii) Calculating the distance covered in 5 seconds:
To find the distance covered, we can use one of the kinematic equations for uniformly accelerated motion. We know:
- Initial speed, [tex]\( v_i \)[/tex]: [tex]\(0 \, \text{m/s}\)[/tex]
- Acceleration, [tex]\( a \)[/tex]: [tex]\(2 \, \text{m/s}^2\)[/tex]
- Time, [tex]\( t \)[/tex]: [tex]\(5 \, \text{s}\)[/tex]
The kinematic equation that relates these quantities is:
[tex]\[ \text{distance} = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2 \][/tex]
Plug in the known values:
[tex]\[ \text{distance} = 0 \cdot 5 \, \text{s} + \frac{1}{2} \cdot 2 \, \text{m/s}^2 \cdot (5 \, \text{s})^2 \][/tex]
Calculate each term:
[tex]\[ = 0 + \frac{1}{2} \cdot 2 \, \text{m/s}^2 \cdot 25 \, \text{s}^2 \][/tex]
[tex]\[ = \frac{1}{2} \cdot 2 \cdot 25 \][/tex]
[tex]\[ = 25 \, \text{m} \][/tex]
So, the distance covered by the vehicle in 5 seconds is [tex]\( 25 \, \text{meters} \)[/tex].
### Summary:
(i) The acceleration of the vehicle is [tex]\( 2 \, \text{m/s}^2 \)[/tex].
(ii) The distance covered in 5 seconds is [tex]\( 25 \, \text{meters} \)[/tex].
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