Question:
The position of a particle moving along $y$-axis is given as $y = t^2 + 2t + 3$. The average acceleration of the particle between $t=3s$ and $t=6s$ (in $ms^{-2}$) is
The position of a particle moving along $y$-axis is given as $y = t^2 + 2t + 3$. The average acceleration of the particle between $t=3s$ and $t=6s$ (in $ms^{-2}$) is
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Differentiate position to get velocity before acceleration.
Updated On: Apr 24, 2026
- $2$
- $5$
- $4$
- $3$
- $6$
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The Correct Option is A
Solution and Explanation
Concept:
• Average acceleration = $\frac{v_2 - v_1}{t_2 - t_1}$
Step 1: Velocity
\[ v = \frac{dy}{dt} = 2t + 2 \]
Step 2: Find velocities
\[ v(3) = 8,\quad v(6) = 14 \]
Step 3: Average acceleration
\[ = \frac{14 - 8}{6 - 3} = \frac{6}{3} = 2 \] Final Conclusion:
\[ = 2 \]
• Average acceleration = $\frac{v_2 - v_1}{t_2 - t_1}$
Step 1: Velocity
\[ v = \frac{dy}{dt} = 2t + 2 \]
Step 2: Find velocities
\[ v(3) = 8,\quad v(6) = 14 \]
Step 3: Average acceleration
\[ = \frac{14 - 8}{6 - 3} = \frac{6}{3} = 2 \] Final Conclusion:
\[ = 2 \]
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