M mindshift New member Joined Jun 12, 2005 Messages 9 Jun 14, 2005 #1 i have a few more questions if you dont mind: when t>0, a particle moves in a line defined by the function s=t+(1/t) . find the position and acceleration of the particle when it comes to rest. ill probably have some more in a bit.
i have a few more questions if you dont mind: when t>0, a particle moves in a line defined by the function s=t+(1/t) . find the position and acceleration of the particle when it comes to rest. ill probably have some more in a bit.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jun 14, 2005 #2 Hello, mindshift! Simple rules to remember: s(t) is the position of the particle. The first derivative, s'(t), is the velocity of the particle: v(t) The second derivative, s"(t) or v'(t), is the acceleration of the particle, a(t). when t>0, a particle moves in a line defined by the function s = t + (1/t) Find the position and acceleration of the particle when it comes to rest. Click to expand... The particle is at rest when its velocity is zero. . . We have: . v(t) .= .s'(t) .= .1 - t<sup>-2</sup> .= .0 . . . Multiply through by t<sup>2</sup>: . t<sup>2</sup> - 1 .= .0 . . ---> . . t = ±1 . . . We are told that t > 0, so t = 1 is the only solution. At t = 1, its position is: . s(1) .= .1 + 1/1 .= .2 The acceleration is: . a(t) .= .v'(t) .= .2t<sup>-3</sup> .= .2/t<sup>3</sup> At t = 1, its acceleration is: . a(1) .= .2/1<sup>3</sup> .= .2
Hello, mindshift! Simple rules to remember: s(t) is the position of the particle. The first derivative, s'(t), is the velocity of the particle: v(t) The second derivative, s"(t) or v'(t), is the acceleration of the particle, a(t). when t>0, a particle moves in a line defined by the function s = t + (1/t) Find the position and acceleration of the particle when it comes to rest. Click to expand... The particle is at rest when its velocity is zero. . . We have: . v(t) .= .s'(t) .= .1 - t<sup>-2</sup> .= .0 . . . Multiply through by t<sup>2</sup>: . t<sup>2</sup> - 1 .= .0 . . ---> . . t = ±1 . . . We are told that t > 0, so t = 1 is the only solution. At t = 1, its position is: . s(1) .= .1 + 1/1 .= .2 The acceleration is: . a(t) .= .v'(t) .= .2t<sup>-3</sup> .= .2/t<sup>3</sup> At t = 1, its acceleration is: . a(1) .= .2/1<sup>3</sup> .= .2
