Question

A particle moves along the `x`-axis in such a way that its position at time `t` is given by `x(t) = (2-t)/(1-t)`. What is the acceleration of the particle at time `t=0`?

Answer 1

`2"ms"^-2`

Explanation:

`a(t)=d/dt[v(t)]=(d^2)/(dt^2)[x(t)]`

`x(t)=(2-t)/(1-t)`

`v(t)=d/dt[(2-t)/(1-t)]=((1-t)d/dt[2-t]-(2-t)d/dt[1-t])/(1-t)^2=((1-t)(-1)-(2-t)(-1))/(1-t)^2=(t-1+2-t)/(1-t)^2=1/(1-t)^2`

`a(t)=d/dt[(1-t)^-2]=-2(1-t)^-3*d/dt[1-t]=-2(1-t)^-3(-1)=2/(1-t)^3`

`a(0)=2/(1-0)^3=2/1^3=2/1=2"ms"^-2`