Question

A particle executes the motion described by x(t) = x₀(1-e^-yt); t>=0 , x₀>0
(a) Where does particle start and with what velocity?
(b) Find maximum and minimum values of x(t), v(t), a(t). Show that x(t) and a(t) increase with time and v(t) decreases with time.

Answer 1

$$\begin{gathered} Equationofmotionofaparticle, \\ x\left(t\right)=x_0\left(1-e^{-yt}\right);t\ge 0,x_0>0---\left(1\right) \\ \left(a\right)particlestartsfromx\left(0\right)=0,att=0 \\ \frac{dx\left(t\right)}{dt}=x_0\left(0-y\times e^{-yt}\right)=x_0\times y\times e^{-yt}=-y\times \left[x\left(t\right)-x_0\right].----\left(2\right) \\ particlestartswithvelocityv=\frac{dx\left(0\right)}{dt}=x_0\times y. \\ Theaccelerationofparticlea=\frac{dv}{dt}=\frac{d^{2}x\left(t\right)}{d{t}^2}=-x_0\left(y^2\times e^{-yt}\right).or \\ a=-x_0{y}^2{e}^{-yt}=y^2\left[x\left(t\right)-x_0\right]---\left(3\right). \\ Fromeqns\left(1\right),\left(2\right)and\left(3\right)wecometoknowthat, \\ \left(b\right)maximumvaluesofx\left(t\right),v\left(t\right)anda\left(t\right)are{x}_o,-x_{0}yand0. \\ theminimumvaluesofx\left(t\right),v\left(t\right)anda\left(t\right)are0,0and-x_0{y}^2. \\ fromeqns\left(1\right),\left(2\right)andthreewecometoknowthatx\left(t\right)anda\left(t\right)increasewithtimewhereasv\left(t\right) \\ decreaseswithtimebynotingtheirsigns. \end{gathered}$$