1) If the position vector of a particle is given as a function of time by the relation \( \hat{x}=\left(\frac{1-2}{1+4}\right) \) ) where \( \stackrel{\rightharpoonup}{c} \) is a constant unit vector, then the algebraic measure of its initial position vector \( = \) \( \qquad \) (a) \( -\frac{1}{2} \) (b) \( \frac{1}{2} \) (c) 1 (d) zero 2) If the position vector of a particle is given as a function of time by the relation \( \bar{x}=\left(\mathrm{e}^{2+1}\right), \overline{\mathrm{c}} \) where \( \overline{\mathrm{c}} \) is a constant unit vector, then the displacement after t seconds \( =( \) \( \qquad \) (a) \( \mathrm{e}^{21} \) (b) \( \mathrm{e}^{2 t-1} \) (C) \( \mathrm{e}^{2 t}+\mathrm{e} \) (d) \( \mathrm{e}\left(\mathrm{e}^{2 \mathrm{t}}-1\right) \) 3) A particle moves in a straight line such that \( v=3 e^{t+2} \), then its starting (initial) velocity is equal to \( \qquad \) (a) 3 (b) e (c) \( 3 \mathrm{e}^{2} \) (d) \( \mathrm{e}^{2} \)

1) We assume the intended position function is \[ \vec{x}(t)=\frac{1-2t}{1+4t}\,\hat{c}, \] where \(\hat{c}\) is a constant unit vector. The initial position is given by evaluating at \(t=0\): \[ \vec{x}(0)=\frac{1-2\cdot0}{1+4\cdot0}\,\hat{c}=\frac{1}{1}\,\hat{c}=\hat{c}. \] Since \(\hat{c}\) is a unit vector the algebraic measure (i.e. its scalar coefficient) is \(1\). Answer: (c) \(1\). 2) We assume the given position function is \[ \vec{x}(t)=e^{2t+1}\,\vec{c}, \] where \(\vec{c}\) is a constant unit vector. The displacement after \(t\) seconds is the change from \(t=0\) to \(t=t\): \[ \Delta \vec{x}=\vec{x}(t)-\vec{x}(0)=e^{2t+1}\,\vec{c}-e^{2\cdot0+1}\,\vec{c}=e^{2t+1}\,\vec{c}-e^1\,\vec{c}. \] Factor out \(e\): \[ \Delta \vec{x}=e\,\left(e^{2t}-1\right)\vec{c}. \] The algebraic measure (scalar magnitude along the direction of \(\vec{c}\)) is \(e\,(e^{2t}-1)\). Answer: (d) \(e\,(e^{2t}-1)\). 3) The velocity is given by \[ v(t)=3e^{t+2}. \] The initial velocity is obtained by evaluating at \(t=0\): \[ v(0)=3e^{0+2}=3e^2. \] Answer: (c) \(3e^2\).