Question

Find velocity vx(t) and coordinate x(t) for a particle of mass m which is subject to the force given by: Fx = F0 e −kt , where F0 and k are known constants. The initial conditions are x = 0 and v = 1 at t = 0.

Answer 1

`v_(x) (t)=1+( F_(0))/(km)(1-e^(-kt))`

`x=t+( F_(0)t)/(km)+( F_(0)t)/(k^(2) m)(e^(-kt)-1)`

Step-by-step explanation:

Given that the force of the particle is,

`F_(x)=F_(0)e^(-kt)`

Now it can be further written as

`m(dv)/(dt)= F_(0)e^(-kt)\\(dv)/(dt)=( F_(0))/(m) e^(-kt)\\dv=( F_(0))/(m)e^(-kt)dt\\ v=( F_(0))/(-km)e^(-kt)+C`

Now the initial conditions are v=1 at t=0.

So,

`1=( F_(0))/(-km)e^(0)+C\\C=1+( F_(0))/(km)`

Now the velocity will become.

`v_(x) (t)=( F_(0))/(-km)e^(-kt)+1+( F_(0))/(km)\\v_(x) (t)=1+( F_(0))/(km)(1-e^(-kt))`

And,

`(dx)/(dt) =1+( F_(0))/(km)(1-e^(-kt))\\dx=(1+( F_(0))/(km)(1-e^(-kt)))dt\\x=t+( F_(0)t)/(km)+( F_(0)t)/(k^(2) m)e^(-kt)+C\\`

And, another initial condition is x=0 at t=0

`0=0+( F_(0)0)/(km)+( F_(0)t)/(k^(2) m)e^(0)+C\\C=-( F_(0)t)/(k^(2) m)`

Now,

`x=t+( F_(0)t)/(km)+( F_(0)t)/(k^(2) m)e^(-kt)+-( F_(0)t)/(k^(2) m)\\x=t+( F_(0)t)/(km)+( F_(0)t)/(k^(2) m)(e^(-kt)-1)`