Question

Find the distance which an object moves in time $t$if it starts from rest and has acceleration$\frac{d^{2}x}{d{t}^2}=g{e}^{-kt}$. Show that for small$t$the result is approximately, and for very large, the speed$\frac{dx}{dt}$is approximately $(1.10)$constant. The constant is called the terminal speed . (This problem corresponds roughly to the motion of a parachutist.)

Short answer

When anobject moves in time $t$,the distance is $x\left(t\right)=\frac{g}{k^2}\left(e^{-kt}-1\right)+\frac{g}{k}t$ if it starts from rest and has acceleration $\frac{d^{2}x}{d{t}^2}=g{e}^{-kt}$.

Step-by-step solution

Given information

It is given that an object moves in time$t$ , and it starts from rest and obtains acceleration $\frac{d^{2}x}{d{t}^2}=g{e}^{-kt}$.

Meaning of differential equation

In mathematics, an equation with only one independent variable and one or more of its derivatives with respect to the variable is referred to as an ordinarydifferentialequation or ODE.In other words, the ODE is a relation with one independent variable x, a real dependent variable y, and several derivatives $y^{\text{'}},y^{\text{'}\text{'}},....,y^n$in relation to $x$.

Find the distance of an object

Find the distance by integration.

$\begin{array}{c}\frac{dx}{dt}=v\left(t\right)\\ =\int g{e}^{-kt}dt\\ =-\frac{g}{k}{e}^{-kt}+C_1\end{array}$

Find the constant.

$\begin{array}{c}v\left(0\right)=0\\ =-\frac{g}{k}+C_1\\ C_1=\frac{g}{k}\end{array}$