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A particle starts from point A and moves along a straight line path with an acceleration given by a = p – qx where p, q are constants and x is distance from point A. The particle stops at point B. The maximum velocity of the particle is

Question

A particle starts from point A and moves along a straight line path with an acceleration given by a = p – qx where p, q are constants and x is distance from point A. The particle stops at point B. The maximum velocity of the particle is

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Solution

The maximum velocity of the particle can be found by using the equations of motion.

Step 1: We know that the acceleration a = p - qx.

Step 2: We can integrate this equation with respect to x to find the velocity v.

∫v dx = ∫(p - qx) dx

This gives us the equation for velocity as a function of x:

v = px - (1/2)qx^2 + C

where C is the constant of integration.

Step 3: The maximum velocity occurs when the acceleration is zero. So, we set a = 0 and solve for x:

0 = p - qx => x = p/q

Step 4: Substitute x = p/q into the velocity equation:

v_max = p*(p/q) - (1/2)q*(p/q)^2 + C => v_max = p^2/q - p^2/2q + C

Step 5: The constant C can be determined by the initial conditions. If the particle starts from rest at point A, then the initial velocity is zero when x = 0. So, C = 0.

Therefore, the maximum velocity of the particle is v_max = p^2/q - p^2/2q = p^2/2q.

This problem has been solved

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