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:
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: \[ \boxed{1.2 \, \text{m/s}} \]
A force of \( (6x^2 - 4x + 3) \, \text{N} \) acts on a body of mass 0.75 kg and displaces it from \( x = 5 \, \text{m} \) to \( x = 2 \, \text{m} \). The work done by the force is