A particle start from rest, with uniform acceleration at time `t`, choose the correct option
A particle start from rest, with uniform acceleration at time `t`, choose the correct option
A_2019_CBT2_E01_016_O01.png)
_2019_CBT2_E01_016_O01.png)
B_2019_CBT2_E01_016_O02.png)
_2019_CBT2_E01_016_O02.png)
C_2019_CBT2_E01_016_O03.png)
_2019_CBT2_E01_016_O03.png)
D_2019_CBT2_E01_016_O04.png)
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Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem step by step, we will analyze the motion of a particle starting from rest with uniform acceleration.
### Step 1: Understand the Given Information
- The particle starts from rest, which means the initial velocity \( u = 0 \).
- The particle has uniform (constant) acceleration, which we will denote as \( a \).
### Step 2: Write the Equation for Velocity
The relationship between acceleration, velocity, and time can be expressed as:
\[
a = \frac{dv}{dt}
\]
Since acceleration is constant, we can integrate this equation. Rearranging gives:
\[
dv = a \, dt
\]
Integrating both sides:
\[
\int dv = \int a \, dt
\]
This results in:
\[
v = at + C
\]
Given that the initial velocity \( u = 0 \) when \( t = 0 \), we find that \( C = 0 \). Therefore, the equation for velocity becomes:
\[
v = at
\]
### Step 3: Write the Equation for Position
The velocity can also be expressed as the derivative of position with respect to time:
\[
v = \frac{dx}{dt}
\]
Substituting our expression for velocity:
\[
\frac{dx}{dt} = at
\]
Now, we integrate this equation:
\[
dx = at \, dt
\]
Integrating both sides gives:
\[
\int dx = \int at \, dt
\]
This results in:
\[
x = \frac{at^2}{2} + C
\]
Again, since the particle starts from the origin (initial position \( x = 0 \) when \( t = 0 \)), we find that \( C = 0 \). Thus, the equation for position becomes:
\[
x = \frac{at^2}{2}
\]
### Step 4: Analyze the Graphs
1. **Acceleration-Time Graph**: Since acceleration is constant, the graph will be a horizontal line at \( a \).
2. **Velocity-Time Graph**: The graph will be a straight line with a slope equal to \( a \) and passing through the origin (0,0).
3. **Position-Time Graph**: The graph will be a parabola opening upwards, represented by the equation \( x = \frac{at^2}{2} \).
### Conclusion
Based on the analysis:
- Acceleration is constant.
- Velocity is a linear function of time, passing through the origin.
- Position is a quadratic function of time, represented by a parabolic graph.
### Final Answer
All options regarding acceleration, velocity, and position are correct.
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