91Ó°ÊÓ

The angular position of the reference line on the object is, $\mathrm{θ}=0.40e^{2t}$.

The radial distance of the point on the object is, $r=4.0 \text{cm}=4.0×{10}^{-2} \text{m}$

The spaceship is taking a turn along a circular path. Hence, we need to use the equations that relate the linear variables (v and a) with the corresponding angular variables ( $\mathrm{Ó\neg }$and $\mathrm{Î\pm }$). The two variables are related through the parameter r, the radius of the path.

Formula

$$\begin{gathered} a_t=\mathrm{Î\pm }r \\ {a}_r=r{\mathrm{Ó\neg }}^2 \\ \mathrm{Î\pm }=\frac{d^2\mathrm{θ}}{d{t}^2} \end{gathered}$$

By using following formula first calculate angular acceleration

$\mathrm{Î\pm }=\frac{d^2\mathrm{θ}}{d{t}^2}$

It is given that, $\mathrm{θ}=0.40e^{2t}$

$\begin{array}{rcl}\mathrm{Ó\neg }& =& \frac{d\mathrm{θ}}{dt}\\ & =& 0.40\left(e^{2t}.\frac{d\left(2t\right)}{dt}\right)\\ \mathrm{Ó\neg }& =& 0.40×2e^{2t}\\ & & \end{array}$

$\begin{array}{rcl}\mathrm{Ó\neg }& =& 0.80e^{2t}\\ \frac{d^2\mathrm{θ}}{d{t}^2}& =& 0.80\left(e^{2t}.\frac{d\left(2t\right)}{dt}\right)\\ & =& 0.80×2e^{2t}\\ & =& 1.60e^{2t}\\ & =& 1.60e^{2t} \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\\ & & \\ & & \end{array}$

Now, the tangential component of the acceleration is given as

$a_t=\mathrm{Î\pm }r$

Substitute all the value in the above equation.

$$\begin{gathered} a_t=\mathrm{Î\pm }r \\ =\left(1.60e^{2t} \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\right)×4.0×{10}^{-2} \text{m} \end{gathered}$$

$\begin{array}{rcl}At& =& 0 \text{s}\\ \mathrm{Î\pm }& =& 1.60e^{2t}\\ & =& 1.60×1\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\\ \mathrm{Î\pm }& =& 1.60 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$

$\begin{array}{rcl}{a}_t& =& \mathrm{Î\pm }r\\ & =& 1.60\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}\text{×}4.0×{10}^{-2} \text{m}}$}\right.\\ & =& 6.4×{10}^{-2}\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\\ & & \end{array}$

The magnitude of the tangential component of acceleration is $6.4×{10}^{-2}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$

Step 3: (b) Calculation of radial component of acceleration

The radial component of the acceleration is calculated as

$a_r=r{\mathrm{Ó\neg }}^2$

$\begin{array}{rcl}Att& =& 0\text{s},\\ \mathrm{Ó\neg }& =& \frac{d\mathrm{θ}}{dt}\\ & =& 0.80e^{2t}\\ & =& 0.80×1 \raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\\ \mathrm{Ó\neg }& =& 0.80\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\\ & & \\ & & \end{array}$

$a_r=r{\mathrm{Ó\neg }}^2$

Substitute all the value in the above equation.

role="math" localid="1660922142356" $\begin{array}{rcl}{a}_r& =& r{\mathrm{Ó\neg }}^2\\ & =& \left(4.0×{10}^{-2}m\right)×{\left(0.80\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2\\ & =& 2.6×{10}^{-2}\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$

The magnitude of the radial component of acceleration is,role="math" localid="1660922172077" $\begin{array}{rcl}& & 2.6×{10}^{-2}\raisebox{1ex}{${\displaystyle \text{m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.\end{array}$ .