IDNLearn.com is designed to help you find the answers you need quickly and easily. Join our knowledgeable community and get detailed, reliable answers to all your questions.
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.
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.
