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 $t=0$ ?

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Answer 1 · 2018-05-17T08:27:13

$2"ms"^-2$

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

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

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

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

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

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