18) A particle moves in a straight line it begins its motios from a fixed point $$ (O) $$ on the straigh line such that the algebraic measure of its acceleration is given in terms of the algebraic measure of its position by the relation $$ a=2 \chi+5 $$. Given that the initial velocity of the particle is $$ 2 \mathrm{~m} / \mathrm{sec} $$., then $$ \chi= $$ $$ \qquad $$ m. at $$ \mathrm{v}=4 \mathrm{~m} / \mathrm{sec} $$. (a) 1 or -6 (b) -1 or 6 (C) -1 (d) 6 19)A particle moves in a straight line from a fixed point on the straight line starting at rest such that $$ \mathrm{a}=\frac{3}{8} x^{2} $$ where a is measured in $$ \mathrm{m} / \mathrm{sec}^{2} $$. and $$ x $$ in metre, then the velocity of the particle when $$ X=2 $$ metres $$ = $$ $$ \qquad $$ $$ \mathrm{m} / \mathrm{sec} $$. (a) $$ \sqrt{2} $$ (b) 2 (c) $$ \sqrt{3} $$ (d) 3 20) A particle moves in a straight line with initial speed $$ 2 \mathrm{~m} / \mathrm{sec} $$. from a fixed point ( 0 ) on a straight line such that $$ \mathrm{a}=\mathrm{e}^{x} $$, then $$ \mathrm{v}^{2}= $$ $$ \qquad $$ (a) $$ 2 e^{x} $$ (b) $$ 2 e^{x}+2 $$ (c) $$ 3 e^{x}-1 $$ (d) $$ 2 e^{x}+1 $$ 21) A particle moves in a straight line with initial speed $$ 8 \mathrm{~m} / \mathrm{sec} $$. from a fixed point ( 0 ) on the straight line such that $$ a=40 e^{-x} $$, then the maximum speed of the particle $$ = $$ $$ \qquad $$ $$ \mathrm{m} / \mathrm{sec} $$. (a) 12 (b) 80 (C) 144 (d) $$ \ln 80 $$

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

### Problem 18
We have the acceleration given by the relation:
$$
a = 2\chi + 5
$$
We know that acceleration $$ a $$ can also be expressed as:
$$
a = \frac{dv}{dt} = v \frac{dv}{d\chi}
$$
Thus, we can set up the equation:
$$
v \frac{dv}{d\chi} = 2\chi + 5
$$
We can separate variables and integrate:
$$
\int v \, dv = \int (2\chi + 5) \, d\chi
$$
This gives us:
$$
\frac{v^2}{2} = \chi^2 + 5\chi + C
$$
Using the initial condition $$ v = 2 $$ when $$ \chi = 0 $$:
$$
\frac{2^2}{2} = 0^2 + 5(0) + C \implies C = 2
$$
Thus, the equation becomes:
$$
\frac{v^2}{2} = \chi^2 + 5\chi + 2
$$
Now, we need to find $$ \chi $$ when $$ v = 4 $$:
$$
\frac{4^2}{2} = \chi^2 + 5\chi + 2
$$
This simplifies to:
$$
8 = \chi^2 + 5\chi + 2 \implies \chi^2 + 5\chi - 6 = 0
$$
Now we can solve this quadratic equation using the quadratic formula:
$$
\chi = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4(1)(-6)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 24}}{2} = \frac{-5 \pm 7}{2}
$$
This gives us:
$$
\chi = 1 \quad \text{or} \quad \chi = -6
$$
Thus, the answer is:
**(a) 1 or -6**

### Problem 19
Given:
$$
a = \frac{3}{8} x^2
$$
We can express acceleration in terms of velocity:
$$
a = v \frac{dv}{dx}
$$
Thus:
$$
v \frac{dv}{dx} = \frac{3}{8} x^2
$$
Separating variables:
$$
\int v \, dv = \int \frac{3}{8} x^2 \, dx
$$
This gives:
$$
\frac{v^2}{2} = \frac{3}{24} x^3 + C
$$
Since the particle starts at rest, when $$ x = 0 $$, $$ v = 0 $$:
$$
0 = 0 + C \implies C = 0
$$
Thus:
$$
\frac{v^2}{2} = \frac{1}{8} x^3
$$
Now, we need to find $$ v $$ when $$ x = 2 $$:
$$
\frac{v^2}{2} = \frac{1}{8} (2^3) = \frac{1}{8} \cdot 8 = 1 \implies v^2 = 2
$$
Thus:
$$
v = \sqrt{2}
$$
The answer is:
**(a) $$ \sqrt{2} $$**

### Problem 20
Given:
$$
a = e^x
$$
We can express acceleration in terms of velocity:
$$
a = v \frac{dv}{dx}
$$
Thus:
$$
v \frac{dv}{dx} = e^x
$$
Separating variables:
$$
\int v \, dv = \int e^x \, dx
$$
This gives:
$$
\frac{v^2}{2} = e^x + C
$$
Using the initial condition $$ v = 2 $$ when $$ x = 0 $$:
$$
\frac{2^2}{2} = e^0 + C \implies 2 = 1 + C \implies C = 1
$$
Thus:
$$
\frac{v^2}{2} = e^x + 1 \implies v^2 = 2e^x + 2
$$
The answer is:
**(b) $$ 2e^x + 2 $$**

### Problem 21
Given:
$$
a = 40 e^{-x}
$$
We can express acceleration in terms of velocity:
$$
a = v \frac{dv}{dx}
$$
Thus:
$$
v \frac{dv}{dx} = 40 e^{-x}
$$
Separating variables:
$$
\int v \, dv = \int 40 e^{-x} \, dx
$$
This gives:
$$
\frac{v^2}{2} = -40 e^{-x} + C
$$
Using the initial condition $$ v = 8 $$ when $$ x = 0 $$:
$$
\frac{8^2}{2} = -40 e^0 + C \implies 32 = -40 + C \implies C = 72
$$
Thus:
$$
\frac{v^2}{2} = -40 e^{-x} + 72 \implies v^2 = -80 e^{-x} + 144
$$
To find the maximum speed, we set $$ \frac{dv}{dx} = 0 $$:
$$
-80 e^{-x} + 144 = 0 \implies e^{-x} = \frac{144}{80} = 1.8
$$
Thus, the maximum speed is:
$$
v^2 = 144 \implies v = 12
$$
The answer is:
**(a) 12**
**Problem 18:** $$ \chi = 1 $$ or $$ \chi = -6 $$ **Answer:** (a) 1 or -6 **Problem 19:** Velocity when $$ x = 2 $$ is $$ \sqrt{2} $$ m/s **Answer:** (a) $$ \sqrt{2} $$ **Problem 20:** $$ v^2 = 2e^x + 2 $$ **Answer:** (b) $$ 2e^x + 2 $$ **Problem 21:** Maximum speed is 12 m/s **Answer:** (a) 12

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