Problem Set 1: Problem 5. Problem: The acceleration of a particle is a = − λ √ v ,where λ is a constant. At time t =0 the particle is located at x =0 and its velocity is v = V . Also, when its position is x = f the particle’s velocity is v = 1 4 V . (a) Determine the constant λ . (b) At what time τ does the particle come to rest? Solution: Since we know the acceleration as a function of velocity, the obvious first step in solving this problem is to begin with the differential equation relating acceleration and velocity, viz., a = v dv dx = ⇒ adx = vdv (a) For the given acceleration, we have − λ √ vdx = vdv = ⇒ dx = − 1 λ √ vdv Then, integrating, we have 8 x o dx = − 1 λ 8 v V √ vdv = ⇒ x = − 2 3 λ v 3 / 2 e e e e v = v v = V Thus, the particle’s position as a function of its velocity is x = − 2 3 λ p v 3 / 2 − V 3 / 2 Q We can now determine the value of
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.