Answer :
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.