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?

May 17, 2018

$2 {\text{ms}}^{-} 2$

Explanation:

$a \left(t\right) = \frac{d}{\mathrm{dt}} \left[v \left(t\right)\right] = \frac{{d}^{2}}{{\mathrm{dt}}^{2}} \left[x \left(t\right)\right]$

$x \left(t\right) = \frac{2 - t}{1 - t}$

$v \left(t\right) = \frac{d}{\mathrm{dt}} \left[\frac{2 - t}{1 - t}\right] = \frac{\left(1 - t\right) \frac{d}{\mathrm{dt}} \left[2 - t\right] - \left(2 - t\right) \frac{d}{\mathrm{dt}} \left[1 - t\right]}{1 - t} ^ 2 = \frac{\left(1 - t\right) \left(- 1\right) - \left(2 - t\right) \left(- 1\right)}{1 - t} ^ 2 = \frac{t - 1 + 2 - t}{1 - t} ^ 2 = \frac{1}{1 - t} ^ 2$

$a \left(t\right) = \frac{d}{\mathrm{dt}} \left[{\left(1 - t\right)}^{-} 2\right] = - 2 {\left(1 - t\right)}^{-} 3 \cdot \frac{d}{\mathrm{dt}} \left[1 - t\right] = - 2 {\left(1 - t\right)}^{-} 3 \left(- 1\right) = \frac{2}{1 - t} ^ 3$

$a \left(0\right) = \frac{2}{1 - 0} ^ 3 = \frac{2}{1} ^ 3 = \frac{2}{1} = 2 {\text{ms}}^{-} 2$