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Sagot :
Let's tackle each part of this question one by one.
### Part (a): Finding the Functional Form of the Acceleration
The given velocity [tex]\( v(t) \)[/tex] is a function of time [tex]\( t \)[/tex]:
[tex]\[ v(t) = 20t - 5t^2 \][/tex]
Acceleration [tex]\( a(t) \)[/tex] is the derivative of velocity with respect to time:
[tex]\[ a(t) = \frac{dv(t)}{dt} \][/tex]
Differentiating [tex]\( v(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ a(t) = \frac{d(20t - 5t^2)}{dt} \][/tex]
[tex]\[ a(t) = 20 - 10t \][/tex]
So, the functional form of the acceleration is:
[tex]\[ a(t) = 20 - 10t \][/tex]
### Part (b): Finding the Instantaneous Velocity at [tex]\( t = 1, 2, 3, \)[/tex] and [tex]\( 5 \)[/tex] Seconds
To find the instantaneous velocities, we need to substitute the given time values into the velocity function [tex]\( v(t) \)[/tex].
1. For [tex]\( t = 1 \)[/tex] second:
[tex]\[ v(1) = 20 \cdot 1 - 5 \cdot 1^2 \][/tex]
[tex]\[ v(1) = 20 - 5 \][/tex]
[tex]\[ v(1) = 15 \, \text{m/s} \][/tex]
2. For [tex]\( t = 2 \)[/tex] seconds:
[tex]\[ v(2) = 20 \cdot 2 - 5 \cdot 2^2 \][/tex]
[tex]\[ v(2) = 40 - 20 \][/tex]
[tex]\[ v(2) = 20 \, \text{m/s} \][/tex]
3. For [tex]\( t = 3 \)[/tex] seconds:
[tex]\[ v(3) = 20 \cdot 3 - 5 \cdot 3^2 \][/tex]
[tex]\[ v(3) = 60 - 45 \][/tex]
[tex]\[ v(3) = 15 \, \text{m/s} \][/tex]
4. For [tex]\( t = 5 \)[/tex] seconds:
[tex]\[ v(5) = 20 \cdot 5 - 5 \cdot 5^2 \][/tex]
[tex]\[ v(5) = 100 - 125 \][/tex]
[tex]\[ v(5) = -25 \, \text{m/s} \][/tex]
### Part (c): Finding the Instantaneous Acceleration at [tex]\( t = 1, 2, 3, \)[/tex] and [tex]\( 5 \)[/tex] Seconds
To find the instantaneous accelerations, we need to substitute the given time values into the acceleration function [tex]\( a(t) \)[/tex].
1. For [tex]\( t = 1 \)[/tex] second:
[tex]\[ a(1) = 20 - 10 \cdot 1 \][/tex]
[tex]\[ a(1) = 20 - 10 \][/tex]
[tex]\[ a(1) = 10 \, \text{m/s}^2 \][/tex]
2. For [tex]\( t = 2 \)[/tex] seconds:
[tex]\[ a(2) = 20 - 10 \cdot 2 \][/tex]
[tex]\[ a(2) = 20 - 20 \][/tex]
[tex]\[ a(2) = 0 \, \text{m/s}^2 \][/tex]
3. For [tex]\( t = 3 \)[/tex] seconds:
[tex]\[ a(3) = 20 - 10 \cdot 3 \][/tex]
[tex]\[ a(3) = 20 - 30 \][/tex]
[tex]\[ a(3) = -10 \, \text{m/s}^2 \][/tex]
4. For [tex]\( t = 5 \)[/tex] seconds:
[tex]\[ a(5) = 20 - 10 \cdot 5 \][/tex]
[tex]\[ a(5) = 20 - 50 \][/tex]
[tex]\[ a(5) = -30 \, \text{m/s}^2 \][/tex]
### Part (d): Interpretation of Results in Terms of the Directions of Acceleration and Velocity
To interpret the results, we need to compare the signs of velocity and acceleration at different time points:
1. At [tex]\( t = 1 \)[/tex] second:
- Velocity [tex]\( v(1) = 15 \, \text{m/s} \)[/tex]
- Acceleration [tex]\( a(1) = 10 \, \text{m/s}^2 \)[/tex]
Both velocity and acceleration are positive, meaning the object is speeding up.
2. At [tex]\( t = 2 \)[/tex] seconds:
- Velocity [tex]\( v(2) = 20 \, \text{m/s} \)[/tex]
- Acceleration [tex]\( a(2) = 0 \, \text{m/s}^2 \)[/tex]
The object is moving with a constant velocity because the acceleration is zero.
3. At [tex]\( t = 3 \)[/tex] seconds:
- Velocity [tex]\( v(3) = 15 \, \text{m/s} \)[/tex]
- Acceleration [tex]\( a(3) = -10 \, \text{m/s}^2 \)[/tex]
Velocity is positive, but acceleration is negative, indicating the object is slowing down.
4. At [tex]\( t = 5 \)[/tex] seconds:
- Velocity [tex]\( v(5) = -25 \, \text{m/s} \)[/tex]
- Acceleration [tex]\( a(5) = -30 \, \text{m/s}^2 \)[/tex]
Both velocity and acceleration are negative, meaning the object is speeding up in the negative direction.
Summary Interpretation:
- When acceleration and velocity have the same sign, the object is speeding up.
- When acceleration and velocity have opposite signs, the object is slowing down.
### Part (a): Finding the Functional Form of the Acceleration
The given velocity [tex]\( v(t) \)[/tex] is a function of time [tex]\( t \)[/tex]:
[tex]\[ v(t) = 20t - 5t^2 \][/tex]
Acceleration [tex]\( a(t) \)[/tex] is the derivative of velocity with respect to time:
[tex]\[ a(t) = \frac{dv(t)}{dt} \][/tex]
Differentiating [tex]\( v(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[ a(t) = \frac{d(20t - 5t^2)}{dt} \][/tex]
[tex]\[ a(t) = 20 - 10t \][/tex]
So, the functional form of the acceleration is:
[tex]\[ a(t) = 20 - 10t \][/tex]
### Part (b): Finding the Instantaneous Velocity at [tex]\( t = 1, 2, 3, \)[/tex] and [tex]\( 5 \)[/tex] Seconds
To find the instantaneous velocities, we need to substitute the given time values into the velocity function [tex]\( v(t) \)[/tex].
1. For [tex]\( t = 1 \)[/tex] second:
[tex]\[ v(1) = 20 \cdot 1 - 5 \cdot 1^2 \][/tex]
[tex]\[ v(1) = 20 - 5 \][/tex]
[tex]\[ v(1) = 15 \, \text{m/s} \][/tex]
2. For [tex]\( t = 2 \)[/tex] seconds:
[tex]\[ v(2) = 20 \cdot 2 - 5 \cdot 2^2 \][/tex]
[tex]\[ v(2) = 40 - 20 \][/tex]
[tex]\[ v(2) = 20 \, \text{m/s} \][/tex]
3. For [tex]\( t = 3 \)[/tex] seconds:
[tex]\[ v(3) = 20 \cdot 3 - 5 \cdot 3^2 \][/tex]
[tex]\[ v(3) = 60 - 45 \][/tex]
[tex]\[ v(3) = 15 \, \text{m/s} \][/tex]
4. For [tex]\( t = 5 \)[/tex] seconds:
[tex]\[ v(5) = 20 \cdot 5 - 5 \cdot 5^2 \][/tex]
[tex]\[ v(5) = 100 - 125 \][/tex]
[tex]\[ v(5) = -25 \, \text{m/s} \][/tex]
### Part (c): Finding the Instantaneous Acceleration at [tex]\( t = 1, 2, 3, \)[/tex] and [tex]\( 5 \)[/tex] Seconds
To find the instantaneous accelerations, we need to substitute the given time values into the acceleration function [tex]\( a(t) \)[/tex].
1. For [tex]\( t = 1 \)[/tex] second:
[tex]\[ a(1) = 20 - 10 \cdot 1 \][/tex]
[tex]\[ a(1) = 20 - 10 \][/tex]
[tex]\[ a(1) = 10 \, \text{m/s}^2 \][/tex]
2. For [tex]\( t = 2 \)[/tex] seconds:
[tex]\[ a(2) = 20 - 10 \cdot 2 \][/tex]
[tex]\[ a(2) = 20 - 20 \][/tex]
[tex]\[ a(2) = 0 \, \text{m/s}^2 \][/tex]
3. For [tex]\( t = 3 \)[/tex] seconds:
[tex]\[ a(3) = 20 - 10 \cdot 3 \][/tex]
[tex]\[ a(3) = 20 - 30 \][/tex]
[tex]\[ a(3) = -10 \, \text{m/s}^2 \][/tex]
4. For [tex]\( t = 5 \)[/tex] seconds:
[tex]\[ a(5) = 20 - 10 \cdot 5 \][/tex]
[tex]\[ a(5) = 20 - 50 \][/tex]
[tex]\[ a(5) = -30 \, \text{m/s}^2 \][/tex]
### Part (d): Interpretation of Results in Terms of the Directions of Acceleration and Velocity
To interpret the results, we need to compare the signs of velocity and acceleration at different time points:
1. At [tex]\( t = 1 \)[/tex] second:
- Velocity [tex]\( v(1) = 15 \, \text{m/s} \)[/tex]
- Acceleration [tex]\( a(1) = 10 \, \text{m/s}^2 \)[/tex]
Both velocity and acceleration are positive, meaning the object is speeding up.
2. At [tex]\( t = 2 \)[/tex] seconds:
- Velocity [tex]\( v(2) = 20 \, \text{m/s} \)[/tex]
- Acceleration [tex]\( a(2) = 0 \, \text{m/s}^2 \)[/tex]
The object is moving with a constant velocity because the acceleration is zero.
3. At [tex]\( t = 3 \)[/tex] seconds:
- Velocity [tex]\( v(3) = 15 \, \text{m/s} \)[/tex]
- Acceleration [tex]\( a(3) = -10 \, \text{m/s}^2 \)[/tex]
Velocity is positive, but acceleration is negative, indicating the object is slowing down.
4. At [tex]\( t = 5 \)[/tex] seconds:
- Velocity [tex]\( v(5) = -25 \, \text{m/s} \)[/tex]
- Acceleration [tex]\( a(5) = -30 \, \text{m/s}^2 \)[/tex]
Both velocity and acceleration are negative, meaning the object is speeding up in the negative direction.
Summary Interpretation:
- When acceleration and velocity have the same sign, the object is speeding up.
- When acceleration and velocity have opposite signs, the object is slowing down.
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