Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Sure! Let's tackle each part of the question step-by-step.
### Problem 2a: Distance traveled by the car
The car starts from rest (initial velocity [tex]\( u = 0 \)[/tex] m/s) and accelerates at [tex]\( 1300 \)[/tex] m/s[tex]\(^2\)[/tex] until it reaches a final velocity of [tex]\( 65 \)[/tex] m/s.
#### To find the time taken to reach the final velocity:
Using the equation of motion:
[tex]\[ v = u + at \][/tex]
where:
- [tex]\( v \)[/tex] is the final velocity ([tex]\( 65 \)[/tex] m/s),
- [tex]\( u \)[/tex] is the initial velocity ([tex]\( 0 \)[/tex] m/s),
- [tex]\( a \)[/tex] is the acceleration ([tex]\( 1300 \)[/tex] m/s[tex]\(^2\)[/tex]),
- [tex]\( t \)[/tex] is the time taken to reach the final velocity.
Rearrange the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{v - u}{a} \][/tex]
[tex]\[ t = \frac{65 - 0}{1300} \][/tex]
[tex]\[ t = \frac{65}{1300} \][/tex]
[tex]\[ t = 0.05 \, \text{s} \][/tex]
So, the car takes [tex]\( 0.05 \)[/tex] seconds to reach the final velocity.
#### To find the distance traveled in this time:
Using the equation of motion:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
where:
- [tex]\( s \)[/tex] is the distance traveled,
- [tex]\( u \)[/tex] is the initial velocity ([tex]\( 0 \)[/tex] m/s),
- [tex]\( a \)[/tex] is the acceleration ([tex]\( 1300 \)[/tex] m/s[tex]\(^2\)[/tex]),
- [tex]\( t \)[/tex] is the time taken ([tex]\( 0.05 \)[/tex] s).
Plugging in the values:
[tex]\[ s = 0 \cdot 0.05 + \frac{1}{2}(1300)(0.05)^2 \][/tex]
[tex]\[ s = 0 + \frac{1}{2}(1300)(0.0025) \][/tex]
[tex]\[ s = 0 + \frac{1300 \cdot 0.0025}{2} \][/tex]
[tex]\[ s = 0 + 1.625 \][/tex]
[tex]\[ s = 1.625 \, \text{m} \][/tex]
Therefore, the distance traveled by the car during the first [tex]\( 0.05 \)[/tex] seconds is [tex]\( 1.625 \)[/tex] meters.
### Problem 3a: Velocity of a particle in rectilinear motion
The velocity of a particle is given by the equation:
[tex]\[ v = 7 + \frac{T}{10} \][/tex]
We need to determine the velocity at a specific [tex]\( T \)[/tex]. Since the problem statement is not complete and the specific [tex]\( T \)[/tex] is not given, let's assume the general form:
If [tex]\( T = 0 \)[/tex]:
[tex]\[ v = 7 + \frac{0}{10} \][/tex]
[tex]\[ v = 7 + 0 \][/tex]
[tex]\[ v = 7 \, \text{m/s} \][/tex]
Therefore, the velocity of the particle at [tex]\( T = 0 \)[/tex] seconds is [tex]\( 7 \)[/tex] m/s.
Please provide any additional [tex]\( T \)[/tex] values if you need the velocity at different times.
### Problem 2a: Distance traveled by the car
The car starts from rest (initial velocity [tex]\( u = 0 \)[/tex] m/s) and accelerates at [tex]\( 1300 \)[/tex] m/s[tex]\(^2\)[/tex] until it reaches a final velocity of [tex]\( 65 \)[/tex] m/s.
#### To find the time taken to reach the final velocity:
Using the equation of motion:
[tex]\[ v = u + at \][/tex]
where:
- [tex]\( v \)[/tex] is the final velocity ([tex]\( 65 \)[/tex] m/s),
- [tex]\( u \)[/tex] is the initial velocity ([tex]\( 0 \)[/tex] m/s),
- [tex]\( a \)[/tex] is the acceleration ([tex]\( 1300 \)[/tex] m/s[tex]\(^2\)[/tex]),
- [tex]\( t \)[/tex] is the time taken to reach the final velocity.
Rearrange the equation to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{v - u}{a} \][/tex]
[tex]\[ t = \frac{65 - 0}{1300} \][/tex]
[tex]\[ t = \frac{65}{1300} \][/tex]
[tex]\[ t = 0.05 \, \text{s} \][/tex]
So, the car takes [tex]\( 0.05 \)[/tex] seconds to reach the final velocity.
#### To find the distance traveled in this time:
Using the equation of motion:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
where:
- [tex]\( s \)[/tex] is the distance traveled,
- [tex]\( u \)[/tex] is the initial velocity ([tex]\( 0 \)[/tex] m/s),
- [tex]\( a \)[/tex] is the acceleration ([tex]\( 1300 \)[/tex] m/s[tex]\(^2\)[/tex]),
- [tex]\( t \)[/tex] is the time taken ([tex]\( 0.05 \)[/tex] s).
Plugging in the values:
[tex]\[ s = 0 \cdot 0.05 + \frac{1}{2}(1300)(0.05)^2 \][/tex]
[tex]\[ s = 0 + \frac{1}{2}(1300)(0.0025) \][/tex]
[tex]\[ s = 0 + \frac{1300 \cdot 0.0025}{2} \][/tex]
[tex]\[ s = 0 + 1.625 \][/tex]
[tex]\[ s = 1.625 \, \text{m} \][/tex]
Therefore, the distance traveled by the car during the first [tex]\( 0.05 \)[/tex] seconds is [tex]\( 1.625 \)[/tex] meters.
### Problem 3a: Velocity of a particle in rectilinear motion
The velocity of a particle is given by the equation:
[tex]\[ v = 7 + \frac{T}{10} \][/tex]
We need to determine the velocity at a specific [tex]\( T \)[/tex]. Since the problem statement is not complete and the specific [tex]\( T \)[/tex] is not given, let's assume the general form:
If [tex]\( T = 0 \)[/tex]:
[tex]\[ v = 7 + \frac{0}{10} \][/tex]
[tex]\[ v = 7 + 0 \][/tex]
[tex]\[ v = 7 \, \text{m/s} \][/tex]
Therefore, the velocity of the particle at [tex]\( T = 0 \)[/tex] seconds is [tex]\( 7 \)[/tex] m/s.
Please provide any additional [tex]\( T \)[/tex] values if you need the velocity at different times.
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.