Question:
A velocity-time graph is drawn for two different objects. They make \( 30^{\circ} \) and \( 45^{\circ} \) with the time axis. Then the ratio of their accelerations, \( a_1: a_2 \) is
A velocity-time graph is drawn for two different objects. They make \( 30^{\circ} \) and \( 45^{\circ} \) with the time axis. Then the ratio of their accelerations, \( a_1: a_2 \) is
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Always remember that on any kinetic plotting graph:
• Slope of Displacement-Time curve \( = \) Velocity.
• Slope of Velocity-Time curve \( = \) Acceleration.
• Slope of Displacement-Time curve \( = \) Velocity.
• Slope of Velocity-Time curve \( = \) Acceleration.
Updated On: Jun 8, 2026
- 1:2
- 2:3
- \( \sqrt{3}:1 \)
- \( 1:\sqrt{3} \)
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The Correct Option is D
Solution and Explanation
Concept:
The instantaneous acceleration of any moving body represents the time-rate change of its velocity vector. On an orthogonal velocity-time coordinate graph, this is exactly equal to the mathematical slope (\( \tan\theta \)) of the line plot relative to the horizontal axis.
Step 1: Formulating acceleration ratios via line slopes.
Let \( \theta_1 = 30^{\circ} \) and \( \theta_2 = 45^{\circ} \). \[ a_1 = \tan(30^{\circ}) = \frac{1}{\sqrt{3}} \] \[ a_2 = \tan(45^{\circ}) = 1 \]
Step 2: Computing the structural ratio.
\[ \frac{a_1}{a_2} = \frac{\frac{1}{\sqrt{3}}}{1} = \frac{1}{\sqrt{3}} \] Thus, the acceleration ratio is \( 1:\sqrt{3} \).
Step 1: Formulating acceleration ratios via line slopes.
Let \( \theta_1 = 30^{\circ} \) and \( \theta_2 = 45^{\circ} \). \[ a_1 = \tan(30^{\circ}) = \frac{1}{\sqrt{3}} \] \[ a_2 = \tan(45^{\circ}) = 1 \]
Step 2: Computing the structural ratio.
\[ \frac{a_1}{a_2} = \frac{\frac{1}{\sqrt{3}}}{1} = \frac{1}{\sqrt{3}} \] Thus, the acceleration ratio is \( 1:\sqrt{3} \).
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