The displacement of a particle is zero at t=0 and it is x at t=t.It st...
Explanation:
The problem states that the displacement of a particle is zero at t=0 and it is x at t=t. The particle starts moving in the positive x direction with a velocity v that varies as v=√x, where k is a constant. We need to show that the motion is uniformly accelerated.
Definition of Uniformly Accelerated Motion:
Uniformly accelerated motion refers to the motion of an object where the acceleration remains constant throughout the motion. In other words, the change in velocity of the object is constant per unit time.
Derivation:
Let's assume the particle's position at time t is given by x(t). According to the problem, the velocity of the particle varies as v=√x. We can relate the velocity and position using the derivative:
v = dx(t)/dt
Since v=√x, we can rewrite the equation as:
√x = dx(t)/dt
To solve this differential equation, we need to eliminate the square root. We can do this by squaring both sides of the equation:
x = (dx(t)/dt)^2
Differentiating both sides with respect to time gives us:
1 = 2(dx(t)/dt)(d^2x(t)/dt^2)
Simplifying the equation, we get:
d^2x(t)/dt^2 = 1/(2(dx(t)/dt))
This equation shows that the acceleration, given by d^2x(t)/dt^2, is inversely proportional to the derivative of the velocity, dx(t)/dt.
Conclusion:
The derived equation d^2x(t)/dt^2 = 1/(2(dx(t)/dt)) indicates that the acceleration of the particle is not constant, but rather inversely proportional to the derivative of the velocity. Since the acceleration is not constant, the motion is not uniformly accelerated.
Therefore, the statement "the motion is uniformly accelerated" is incorrect based on the given information and the derived equation.
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