Question:
A particle under constant acceleration covers the first half of the total distance in time $t_1$ and remaining in time $t_2$. Then:
A particle under constant acceleration covers the first half of the total distance in time $t_1$ and remaining in time $t_2$. Then:
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With constant acceleration, velocity increases over time, so the particle covers the second half of the distance faster than the first.
Updated On: Jun 10, 2026
- $t_1 = t_2$
- $t_1 < t_2$
- $t_1 > t_2$
- $t_1 = t_2^2$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Kinematics of constant acceleration.
Step 2: Analysis
For a particle starting from rest ($u=0$) under constant acceleration ($a$), the distance $s = \frac{1}{2}at^2$. The first half ($s/2$) is covered in time $t_1 = \sqrt{s/a}$. The total distance ($s$) is covered in time $T = \sqrt{2s/a}$. The second half is covered in $t_2 = T - t_1 = \sqrt{s/a}(\sqrt{2} - 1)$.
Step 3: Conclusion
Since $\sqrt{2} - 1 \approx 0.414$, $t_2 < t_1$ is false; actually, the time taken to cover the same distance increases as velocity increases, so $t_1 > t_2$. *Correction*: Re-evaluating the standard kinematic result for constant acceleration from rest: the time to cover successive equal distances is in ratio $1 : (\sqrt{2}-1) : (\sqrt{3}-\sqrt{2})$. Thus, $t_1 > t_2$.
Final Answer: (C)
Kinematics of constant acceleration.
Step 2: Analysis
For a particle starting from rest ($u=0$) under constant acceleration ($a$), the distance $s = \frac{1}{2}at^2$. The first half ($s/2$) is covered in time $t_1 = \sqrt{s/a}$. The total distance ($s$) is covered in time $T = \sqrt{2s/a}$. The second half is covered in $t_2 = T - t_1 = \sqrt{s/a}(\sqrt{2} - 1)$.
Step 3: Conclusion
Since $\sqrt{2} - 1 \approx 0.414$, $t_2 < t_1$ is false; actually, the time taken to cover the same distance increases as velocity increases, so $t_1 > t_2$. *Correction*: Re-evaluating the standard kinematic result for constant acceleration from rest: the time to cover successive equal distances is in ratio $1 : (\sqrt{2}-1) : (\sqrt{3}-\sqrt{2})$. Thus, $t_1 > t_2$.
Final Answer: (C)
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