Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Connect with a community of experts ready to help you find accurate solutions to your questions quickly and efficiently. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
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].
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.