# A particle moves along a straight line with velocity given by v(t) = 7 - (1.01)^(-t^2) at time t>=0. What is the acceleration of the particle at time t = 3?

Jan 26, 2018

The acceleration is $= 0.055 m {s}^{-} 2$

#### Explanation:

$\text{Reminder}$

Let $y = {a}^{{x}^{2}}$

$\ln y = {x}^{2} \ln a$

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x \ln a$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x y \ln a = 2 x {a}^{{x}^{2}} \ln a$

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

$v \left(t\right) = 7 - {\left(1.01\right)}^{- {t}^{2}} = 7 - {\left(\frac{101}{100}\right)}^{- {t}^{2}}$

$= 7 - {\left(\frac{100}{101}\right)}^{{t}^{2}}$

$= 7 - {100}^{{t}^{2}} / {101}^{{t}^{2}}$

And

$\left({100}^{{t}^{2}}\right) ' = 2 t \cdot {100}^{{t}^{2}} \ln 100$

$\left({101}^{{t}^{2}}\right) ' = 2 t \cdot {101}^{{t}^{2}} \ln 101$

Therefore,

$a \left(t\right) = v ' \left(t\right) = \left(- {100}^{{t}^{2}} / {101}^{{t}^{2}}\right) '$

$= - \frac{\left(2 t \cdot {100}^{{t}^{2}} \ln 100\right) \cdot \left({101}^{{t}^{2}}\right) - \left(2 t \cdot {101}^{{t}^{2}} \ln 101\right) \cdot \left({100}^{{t}^{2}}\right)}{{101}^{{t}^{2}}} ^ 2$

$= \frac{- 2 t \ln 100 \cdot {100}^{{t}^{2}} + 2 t \ln 101 \cdot {100}^{{t}^{2}}}{101} ^ \left({t}^{2}\right)$

$= \frac{2 t \left(\ln 101 - \ln 100\right)}{101} ^ \left({t}^{2}\right) \cdot {100}^{{t}^{2}}$

The acceleration when $t = 3$ is

$a \left(3\right) = \frac{2 \cdot 3 \cdot \left(\ln 101 - \ln 100\right)}{101} ^ \left({3}^{2}\right) \cdot {100}^{{3}^{2}}$

$= 0.055 m {s}^{-} 2$