Question

The acceleration of a particle which moves along the positive \( x \)-axis varies with its position as shown in the figure. If the velocity of the particle is \( 0.8 \, \text{m/s} \) at \( x = 0 \), then its velocity at \( x = 1.4 \, \text{m} \) is:

• A. \( 1.6 \, \text{m/s} \)
• B. \( 1.2 \, \text{m/s} \)
• C. \( 1.4 \, \text{m/s} \)
• D. \( 0.8 \, \text{m/s} \)

Hint:

When working with position and acceleration graphs, calculate the area under the curve to determine the change in velocity.

Answer 1

The Correct Option is B

The velocity of the particle can be found using the work-energy theorem, which relates the change in velocity to the area under the acceleration versus position curve. 
Given that the velocity at \( x = 0 \) is \( 0.8 \, \text{m/s} \), and from the graph, the area under the curve between \( x = 0 \) and \( x = 1.4 \) is the work done, 
which contributes to the change in velocity. 
From the graph, we can estimate the area, which leads to a velocity of \( 1.2 \, \text{m/s} \) at \( x = 1.4 \). 
Thus, the velocity at \( x = 1.4 \, \text{m} \) is: \[ \boxed{1.2 \, \text{m/s}} \]

Answer 2

Given:
- Acceleration \( a(x) \) varies with position \( x \) as shown in the graph.
- Initial velocity at \( x = 0 \) is \( v_0 = 0.8 \, \text{m/s} \).
- Find velocity at \( x = 1.4 \, \text{m} \).

Step 1: Use work-energy principle with acceleration as function of position:
\[ v^2 = v_0^2 + 2 \int_0^x a(x) \, dx \] Calculate the integral \( \int_0^{1.4} a(x) \, dx \) from the graph.

Step 2: The graph shows acceleration values:
- From \( x=0 \) to \( 0.4 \): \( a = 0.4 \, \text{m/s}^2 \)
- From \( 0.4 \) to \( 0.8 \): \( a \) decreases linearly from 0.4 to 0.2
- From \( 0.8 \) to \( 1.4 \): \( a = 0.2 \, \text{m/s}^2 \)

Step 3: Calculate area under acceleration curve (integral of acceleration w.r.t position):
- For \( 0 \le x \le 0.4 \):
\[ \int_0^{0.4} 0.4 \, dx = 0.4 \times 0.4 = 0.16 \]

- For \( 0.4 \le x \le 0.8 \):
Acceleration decreases linearly from 0.4 to 0.2, average acceleration:
\[ \frac{0.4 + 0.2}{2} = 0.3 \] Interval length = 0.4, so area:
\[ 0.3 \times 0.4 = 0.12 \]

- For \( 0.8 \le x \le 1.4 \):
Acceleration is constant at 0.2:
\[ 0.2 \times (1.4 - 0.8) = 0.2 \times 0.6 = 0.12 \]

Step 4: Total integral:
\[ \int_0^{1.4} a(x) dx = 0.16 + 0.12 + 0.12 = 0.4 \]

Step 5: Use the relation for velocity:
\[ v^2 = v_0^2 + 2 \times 0.4 = (0.8)^2 + 0.8 = 0.64 + 0.8 = 1.44 \] \[ v = \sqrt{1.44} = 1.2 \, \text{m/s} \]

Therefore, the velocity at \( x = 1.4 \, \text{m} \) is:
\[ \boxed{1.2 \, \text{m/s}} \]