8)A body moves in a straight line such that its position vector is \( \vec{x}=\left(\frac{3 t-2}{t+1}\right) \vec{c} \), then its displacement is \( \qquad \) (a) \( \vec{s}=\left(\frac{3 t-2}{t+1}\right) \vec{c} \) (b) \( \hat{s}=\left(\frac{3 t}{1+1}\right) \hat{c} \) (c) \( \vec{s}=\left(\frac{5 t}{t+1}\right) \overrightarrow{\mathrm{c}} \) (d) \( \bar{s}=\left(\frac{3 t-2}{t+1}\right) \vec{c} \) 9) A particle moves with speed (v) in a straight line such that its position vector \( \vec{r}=4 t(t+1) \hat{i}+15 t \hat{j} \), then (a) \( v(2)=35, a \) is constant. (b) \( \vee(2)=35, a \) is variable. (c) \( v(2)=25, a \) is constant (d) \( v(2)=25, a \) is variable. 10) The motion equation of a particle is \( s=2 t^{3}-3 t^{2}+15 t-8 \), then its initial speed \( = \). (a) 8 (b) 15 (c) -6 (d) 23 11) If \( x=t^{2}-3 t+2 \), then the particle changes its motion direction when (a) \( t=1, t=2 \) (b) \( t=1 \) (c) \( t=1.5 \) (d) \( t=2 \)

**Problem 8** The position vector is given by \[ \vec{x}(t)=\frac{3t-2}{t+1}\,\vec{c}. \] The displacement of a particle (from its initial position) is defined as the difference between its position at time \( t \) and its initial position at \( t=0 \): \[ \vec{s}(t)=\vec{x}(t)-\vec{x}(0). \] 1. Find the initial position: \[ \vec{x}(0)=\frac{3(0)-2}{0+1}\,\vec{c}=\frac{-2}{1}\,\vec{c}=-2\,\vec{c}. \] 2. Write the displacement: \[ \vec{s}(t)=\frac{3t-2}{t+1}\,\vec{c}-(-2\,\vec{c})=\left(\frac{3t-2}{t+1}+2\right)\vec{c}. \] 3. Simplify the expression: \[ \frac{3t-2}{t+1}+2=\frac{3t-2+2(t+1)}{t+1}=\frac{3t-2+2t+2}{t+1}=\frac{5t}{t+1}. \] Thus, the displacement is \[ \vec{s}(t)=\frac{5t}{t+1}\,\vec{c}. \] The correct answer is **(c)**. --- **Problem 9** The position vector is given by \[ \vec{r}(t)=4t(t+1)\,\hat{i}+15t\,\hat{j}. \] We need to determine the speed at \( t=2 \) and decide whether acceleration is constant or variable. 1. Find the \( x \)-component of the position: \[ x(t)=4t(t+1)=4t^2+4t. \] 2. Differentiate to obtain the \( x \)-component of velocity: \[ v_x(t)=\frac{dx}{dt}=8t+4. \] 3. The \( y \)-component of velocity is: \[ v_y(t)=\frac{dy}{dt}=15. \] 4. At \( t=2 \): \[ v_x(2)=8(2)+4=20 \quad \text{and} \quad v_y(2)=15. \] 5. The speed at \( t=2 \) is: \[ v(2)=\sqrt{(20)^2+(15)^2}=\sqrt{400+225}=\sqrt{625}=25. \] 6. Find the acceleration components: - For \( x \)-component: \[ a_x(t)=\frac{dv_x}{dt}=8, \] - For \( y \)-component: \[ a_y(t)=\frac{dv_y}{dt}=0. \] Since the acceleration components are constant, the acceleration is constant. Thus, the correct answer is **(c)**: \( v(2)=25 \) and acceleration is constant. --- **Problem 10** The motion is given by the equation: \[ s=2t^3-3t^2+15t-8. \] The initial speed is the derivative \( \frac{ds}{dt} \) evaluated at \( t=0 \). 1. Differentiate \( s \) with respect to \( t \): \[ \frac{ds}{dt}=6t^2-6t+15. \] 2. Evaluate at \( t=0 \): \[ \frac{ds}{dt}\Big|_{t=0}=6(0)^2-6(0)+15=15. \] The initial speed is \( 15 \). The correct answer is **(b)**. --- **Problem 11** The position is given by \[ x=t^2-3t+2. \] The particle changes direction when the velocity \( \frac{dx}{dt} \) changes sign. First, compute the velocity: 1. Differentiate \( x \) with respect to \( t \): \[ \frac{dx}{dt}=2t-3. \] 2. Set the velocity equal to zero to find when the change of direction might occur: \[ 2t-3=0\quad\Rightarrow\quad t=\frac{3}{2}=1.5. \] Thus, the particle changes its direction at \( t=1.5 \). The correct answer is **(c)**.