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) $$ t=2 $$

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Let's solve each of the problems step by step.

### Problem 8
The position vector of the body is given by:
$$
\vec{x} = \left(\frac{3t - 2}{t + 1}\right) \vec{c}
$$
The displacement $$\vec{s}$$ is typically defined as the change in position. Since the position vector is already given, we can conclude that the displacement is equal to the position vector itself.

Thus, the correct answer is:
$$
\vec{s} = \left(\frac{3t - 2}{t + 1}\right) \vec{c}
$$
**Answer: (a)**

### Problem 9
The position vector of the particle is given by:
$$
\vec{r} = 4t(t + 1) \hat{i} + 15t \hat{j}
$$
To find the speed $$v$$, we need to differentiate the position vector with respect to time $$t$$:
$$
\vec{v} = \frac{d\vec{r}}{dt}
$$
Calculating the derivatives:
1. For the $$i$$ component:
   $$
   \frac{d}{dt}[4t(t + 1)] = 4(2t + 1) = 8t + 4
   $$
2. For the $$j$$ component:
   $$
   \frac{d}{dt}[15t] = 15
   $$
Thus, the velocity vector is:
$$
\vec{v} = (8t + 4) \hat{i} + 15 \hat{j}
$$
Now, we can find the speed at $$t = 2$$:
$$
v(2) = \sqrt{(8(2) + 4)^2 + 15^2} = \sqrt{(16 + 4)^2 + 15^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25
$$
Next, we need to check if the acceleration $$a$$ is constant. The acceleration is given by:
$$
\vec{a} = \frac{d\vec{v}}{dt} = \left(\frac{d}{dt}(8t + 4)\right) \hat{i} + \left(\frac{d}{dt}(15)\right) \hat{j} = 8 \hat{i} + 0 \hat{j}
$$
Since the acceleration is constant, we conclude:
**Answer: (c)**

### Problem 10
The motion equation of the particle is given by:
$$
s = 2t^3 - 3t^2 + 15t - 8
$$
To find the initial speed, we need to differentiate $$s$$ with respect to $$t$$:
$$
v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 3t^2 + 15t - 8)
$$
Calculating the derivative:
$$
v = 6t^2 - 6t + 15
$$
Now, we find the initial speed by evaluating $$v$$ at $$t = 0$$:
$$
v(0) = 6(0)^2 - 6(0) + 15 = 15
$$
**Answer: (b)**

### Problem 11
The position function is given by:
$$
x = t^2 - 3t + 2
$$
To find when the particle changes its motion direction, we need to find the critical points by differentiating $$x$$:
$$
\frac{dx}{dt} = 2t - 3
$$
Setting the derivative equal to zero to find critical points:
$$
2t - 3 = 0 \implies t = \frac{3}{2} = 1.5
$$
Next, we check the second derivative to confirm the change in direction:
$$
\frac{d^2x}{dt^2} = 2
$$
Since the second derivative is positive, the particle changes direction at $$t = 1.5$$.

Thus, the correct answer is:
**Answer: (c)**

### Summary of Answers
8) (a)  
9) (c)  
10) (b)  
11) (c)
**Answers:** 8) (a) 9) (c) 10) (b) 11) (c)

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