To tackle this problem, we need to analyze the motion of the particle given its acceleration as a function of time. The acceleration is defined as \( a = 2t \) m/s². Since the particle starts from rest at the origin, we can derive both the velocity and the displacement over the time interval from \( t = 0 \) to \( t = 3 \) seconds. Let's break this down step by step.
Step 1: Finding Velocity
The acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \). Therefore, we can express the relationship as:
\( a = \frac{dv}{dt} = 2t \)
To find the velocity, we integrate the acceleration with respect to time:
\( v(t) = \int a \, dt = \int 2t \, dt = t^2 + C \)
Since the particle starts from rest at \( t = 0 \), we have \( v(0) = 0 \), which gives us \( C = 0 \). Thus, the velocity function becomes:
\( v(t) = t^2 \)
Step 2: Calculating Average Velocity \( v_1 \)
The time average velocity \( v_1 \) over the interval from \( t = 0 \) to \( t = 3 \) seconds is given by:
\( v_1 = \frac{1}{T} \int_0^T v(t) \, dt \)
Substituting \( T = 3 \) seconds:
\( v_1 = \frac{1}{3} \int_0^3 t^2 \, dt \)
Now, we compute the integral:
\( \int_0^3 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^3 = \frac{27}{3} - 0 = 9 \)
Thus, we find:
\( v_1 = \frac{9}{3} = 3 \, \text{m/s} \)
Step 3: Finding Displacement
Next, we need to calculate the displacement \( s \) over the same time interval. Displacement is given by the integral of velocity:
\( s = \int_0^3 v(t) \, dt = \int_0^3 t^2 \, dt = 9 \, \text{m} \, \text{(as calculated earlier)} \)
Step 4: Calculating Distance Average Velocity \( v_2 \)
The distance average velocity \( v_2 \) is defined as the total displacement divided by the total time:
\( v_2 = \frac{s}{T} = \frac{9}{3} = 3 \, \text{m/s} \)
Step 5: Confirming the Results
Now, let's summarize our findings:
- Time average velocity \( v_1 = 3 \, \text{m/s} \)
- Distance average velocity \( v_2 = 3 \, \text{m/s} \)
However, we need to check the options provided in the question. The average velocities calculated do not match the options given. Let's re-evaluate the average velocity calculations:
For \( v_2 \), we need to consider the average of the instantaneous velocities over the interval:
\( v_2 = \frac{s}{T} = \frac{9}{3} = 3 \, \text{m/s} \)
It appears that the options provided may not align with our calculations based on the given acceleration function. Therefore, none of the options (A, B, C, D) are correct based on the calculations we performed. If there are any additional details or constraints in the problem, please share them for further analysis.
