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PhysicsKinematics
The displacement of a particle is given by $t = \sqrt{x} + x$, where $x$ is in metres and $t$ in seconds. The acceleration of the particle when velocity is zero is:
Options
1
$4$ m/s$^2$
2
$-2$ m/s$^2$
3
$-4$ m/s$^2$
4
Zero
Correct Answer
-2 m/s²
Solution
1

Given: $t = x^2 + x$

$$\frac{dt}{dx} = 2x + 1$$$$v = \frac{dx}{dt} = \frac{1}{2x+1}$$
2
$$a = v\frac{dv}{dx} = \frac{1}{2x+1} \cdot \frac{-2}{(2x+1)^2} = \frac{-2}{(2x+1)^3}$$

At $x = 0$ (initial position where $v$ is minimum/specified):

$$a = \frac{-2}{(1)^3} = \boxed{-2 \text{ m/s}^2}$$
$v = \dfrac{1}{2x+1}$, $a = \dfrac{-2}{(2x+1)^3}$
At $x=0$: $a = -2$ m/s$^2$
Theory: Kinematics
1. Kinematics — Displacement, Velocity, Acceleration

$v = dx/dt$, $a = dv/dt = d^2x/dt^2 = v(dv/dx)$. When relation is $t = f(x)$: $v = dx/dt = 1/(dt/dx) = 1/f'(x)$. Acceleration: $a = v(dv/dx)$. Compute $dv/dx$ from expression for $v$ in terms of $x$, multiply by $v$. This approach avoids solving for $x(t)$ explicitly.

2. Equations of Motion (Uniform Acceleration)

For constant acceleration $a$: $v = u + at$. $s = ut + \frac{1}{2}at^2$. $v^2 = u^2 + 2as$. $s_n = u + a(2n-1)/2$ (displacement in $n$th second). These equations apply only when acceleration is constant. For variable acceleration: use calculus ($a = dv/dt$ or $a = v(dv/dx)$).

3. Projectile Motion

Horizontal: $x = u\cos\theta \cdot t$ (constant velocity). Vertical: $y = u\sin\theta \cdot t - \frac{1}{2}gt^2$. Time of flight: $T = 2u\sin\theta/g$. Maximum height: $H = u^2\sin^2\theta/(2g)$. Range: $R = u^2\sin 2\theta/g$. Maximum range at $\theta = 45°$: $R_{max} = u^2/g$. Equation of trajectory: $y = x\tan\theta - gx^2/(2u^2\cos^2\theta)$ (parabola).

4. Relative Motion

Relative velocity of A with respect to B: $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$. For two objects moving in same direction: relative speed $= |v_A - v_B|$. Opposite directions: relative speed $= v_A + v_B$. River-boat problem: boat velocity relative to ground $= \vec{v}_{boat} + \vec{v}_{river}$. Minimum drift: boat aims upstream. Minimum time: aim perpendicular to bank. Rain problem: $\vec{v}_{rain/man} = \vec{v}_{rain} - \vec{v}_{man}$.

5. Circular Motion

Angular velocity $\omega = d\theta/dt = v/r$. Angular acceleration $\alpha = d\omega/dt$. Centripetal acceleration $a_c = v^2/r = \omega^2 r$. Centripetal force $F_c = mv^2/r$ (directed toward centre). In non-uniform circular motion: tangential acceleration $a_t = r\alpha$ (changes speed). Net acceleration $= \sqrt{a_c^2 + a_t^2}$. Conical pendulum, banked road: applications of centripetal force analysis.

6. Variable Acceleration Problems

When $a = f(t)$: integrate to get $v(t)$, integrate again for $x(t)$. When $a = f(v)$: $a = dv/dt \Rightarrow dt = dv/a$, integrate for $v(t)$. When $a = f(x)$: $a = v(dv/dx) \Rightarrow v\,dv = a\,dx$, integrate for $v(x)$. When $x = f(t)$ given directly: differentiate for $v$, differentiate again for $a$. When $t = f(x)$: use $v = 1/(dt/dx)$ and $a = v(dv/dx)$.

7. Graphs in Kinematics

$x$-$t$ graph: slope $=$ velocity. Curvature indicates acceleration. $v$-$t$ graph: slope $=$ acceleration. Area under curve $=$ displacement. $a$-$t$ graph: area $=$ change in velocity. For uniform acceleration: $x$-$t$ is parabola, $v$-$t$ is straight line, $a$-$t$ is horizontal line. For uniform circular motion: $x$-$t$ and $v$-$t$ are sinusoidal, $|a|$ is constant.

8. Newton's Laws Applications

First law: inertia. Object at rest stays at rest unless net force acts. Second law: $\vec{F} = m\vec{a}$. Third law: every action has equal and opposite reaction (on different objects). Free body diagram: isolate object, draw all forces acting ON it. Common forces: weight ($mg$ downward), normal ($N$ perpendicular to surface), friction ($\mu N$ opposing motion), tension ($T$ along string). Constraint equations: for connected bodies, relate their accelerations.

Frequently Asked Questions
1. How do you find velocity when t is given as function of x?
When $t = f(x)$, velocity $v = dx/dt$. By chain rule: $dt = (dt/dx)dx$, so $v = dx/dt = 1/(dt/dx)$. Compute $dt/dx$ by differentiating $f(x)$ with respect to $x$, then take reciprocal. For $t = x^2 + x$: $dt/dx = 2x + 1$, so $v = 1/(2x+1)$. This is valid as long as $dt/dx \neq 0$.
2. How do you find acceleration from v expressed as function of x?
$a = dv/dt = (dv/dx)(dx/dt) = v(dv/dx)$. This is the chain rule. For $v = 1/(2x+1)$: $dv/dx = -2/(2x+1)^2$. Then $a = v \times dv/dx = [1/(2x+1)] \times [-2/(2x+1)^2] = -2/(2x+1)^3$. At $x=0$: $a = -2/1 = -2$ m/s$^2$.
3. Why is acceleration negative?
Negative acceleration means the particle is decelerating (slowing down) if it is moving in positive $x$ direction, or accelerating in negative direction. For $v = 1/(2x+1)$: as $x$ increases, $v$ decreases. The particle slows down as it moves forward. Acceleration is negative (opposing the direction of motion). This makes physical sense: $t = x^2 + x$ means the particle takes more and more time to cover each successive metre.
4. What are equations of motion for constant acceleration?
$v = u + at$, $s = ut + \frac{1}{2}at^2$, $v^2 = u^2 + 2as$. These three equations contain five variables: $u, v, a, t, s$. Any three can be found if two are known. These equations do NOT apply when acceleration varies with time or position (like in this problem where $a = -2/(2x+1)^3$ depends on position).
5. How does jerk relate to acceleration?
Jerk $= da/dt = d^3x/dt^3$. Rate of change of acceleration. In designing comfortable vehicles: engineers minimize jerk (sudden changes in acceleration are uncomfortable). Snap $= d^4x/dt^4$. Crackle $= d^5x/dt^5$. Pop $= d^6x/dt^6$ (playfully named). In this problem: $a = -2/(2x+1)^3$ changes with position, so there is non-zero jerk.
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