A particle at rest starts from the origin with a constant acceleration a that makes an angle 60° with the positive y-axis. If its displacement along y-axis is 10m in time 2s,then the magnitude of a is:
10ms-2
4ms-2
8ms-2
15ms-2
20ms-2
The Correct Option is A
Approach Solution - 1
Given:
- Particle starts from rest at the origin.
- Constant acceleration \( \vec{a} \) makes a \( 60^\circ \) angle with the positive y-axis.
- Displacement along y-axis is 10 m in 2 seconds.
Step 1: Resolve Acceleration into Components
The acceleration \( \vec{a} \) can be resolved into:
\[ a_y = a \cos 60^\circ = \frac{a}{2} \]
\[ a_x = a \sin 60^\circ = \frac{a\sqrt{3}}{2} \]
Step 2: Apply Kinematic Equation for y-Displacement
Since the particle starts from rest, its initial velocity \( u_y = 0 \). The displacement along the y-axis is given by:
\[ y = u_y t + \frac{1}{2} a_y t^2 \]
Substitute \( y = 10 \, \text{m} \), \( t = 2 \, \text{s} \), and \( a_y = \frac{a}{2} \):
\[ 10 = 0 + \frac{1}{2} \left( \frac{a}{2} \right) (2)^2 \]
\[ 10 = \frac{a}{4} \times 4 \]
\[ 10 = a \]
Conclusion:
The magnitude of the acceleration \( \vec{a} \) is 10 ms\(^{-2}\).
Answer: \(\boxed{A}\)
Approach Solution -2
Step 1: Decompose the Acceleration Vector
The acceleration \( \vec{a} \) can be expressed in terms of its components along the x- and y-axes:
\[ a_y = a \cos 60^\circ = \frac{a}{2}, \quad a_x = a \sin 60^\circ = \frac{a\sqrt{3}}{2}. \]
Here, \( a_y \) represents the component of acceleration along the y-axis, which governs the motion in the vertical direction.
Step 2: Use Kinematic Equation for y-Displacement
The particle starts from rest (\( u_y = 0 \)), so its displacement along the y-axis after \( t = 2 \, \text{s} \) is governed by the kinematic equation:
\[ y = u_y t + \frac{1}{2} a_y t^2. \]
Substituting the known values (\( y = 10 \, \text{m} \), \( t = 2 \, \text{s} \), and \( a_y = \frac{a}{2} \)):
\[ 10 = 0 + \frac{1}{2} \left( \frac{a}{2} \right) (2)^2. \]
Simplify the equation:
\[ 10 = \frac{1}{2} \cdot \frac{a}{2} \cdot 4, \]
\[ 10 = \frac{a}{4} \cdot 4, \]
\[ 10 = a. \]
Conclusion:
The magnitude of the acceleration \( \vec{a} \) is \( \boxed{10 \, \text{ms}^{-2}} \).
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Concepts Used:
Acceleration
In the real world, everything is always in motion. Objects move at a variable or a constant speed. When someone steps on the accelerator or applies brakes on a car, the speed of the car increases or decreases and the direction of the car changes. In physics, these changes in velocity or directional magnitude of a moving object are represented by acceleration.
