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
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.
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