iCON Education - Kinematics-1D Advanced Test

Concept: Velocity $v$ is the first derivative of displacement $s$ with respect to time $t$, given by $v = \frac{ds}{dt}$. Acceleration $a$ is the derivative of velocity with respect to time $t$, given by $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$.

If $s = 2t^3 + 3t^2 + 2t + 8$, then the time at which acceleration is zero, is :

$t = \frac{1}{2\sqrt{2}}$

Concept: Displacement $s$ is obtained by integrating velocity $v$ with respect to time $t$ over the given time interval, $s = \int_{t_1}^{t_2} v \, dt$.

Velocity of a particle varies with time as $v = 4t$. The displacement of particle between $t = 2$ to $t = 4\text{ sec}$, is :

Concept: Distance travelled in the $n^{\text{th}}$ second is the integral of velocity $v$ with respect to time $t$ from $t = n - 1$ to $t = n$, given by $\text{Distance} = \int_{n-1}^{n} v \, dt$, provided velocity does not change direction in that interval.

A point mass moves with velocity $v = (5t - t^2)\text{ ms}^{-1}$ in a straight line. Find the distance travelled (i.e. $\int v \, dt$) in fourth second.

$\frac{31}{6}\text{ m}$

$\frac{29}{6}\text{ m}$

$\frac{37}{6}\text{ m}$

Concept: When acceleration $a$ is expressed as a function of position $x$, acceleration is related to velocity $v$ by $a = v \frac{dv}{dx}$. The stopping distance is found by integrating this differential equation from initial velocity $v = v_0$ (at $x = 0$) to final velocity $v = 0$ (at stopping distance $x$).

A particle is projected with velocity $v_0$ along x-axis. The deceleration on the particle is proportional to the square of the distance from the origin i.e., $a = -\alpha x^2$. The distance at which the particle stops is:

$\left(\frac{3v_0}{2\alpha}\right)^{\frac{1}{3}}$

$\left(\frac{3v_0}{2\alpha}\right)^3$

$\sqrt{\frac{3v_0^2}{2\alpha}}$

$\left(\frac{3v_0^2}{2\alpha}\right)^{\frac{1}{3}}$

Concept: When acceleration $a$ is given as a function of position $x$, it can be expressed as $a = v \frac{dv}{dx}$. The distance covered is found by integrating this relation between the initial state ($x = 0$, $v = v_0$) and final state ($x$, $v = 2v_0$).

The acceleration vector along x-axis of a particle having initial speed $v_0$ changes with distance as $a = \sqrt{x}$. The distance covered by the particle, when its speed becomes twice that of initial speed is:

$\left(\frac{9}{4} v_0\right)^{\frac{4}{3}}$

$\left(\frac{3}{2} v_0\right)^{\frac{4}{3}}$

$\left(\frac{2}{3} v_0\right)^{\frac{4}{3}}$

Concept: Velocity $v$ is the first derivative of position $x$ with respect to time $t$, given by $v = \frac{dx}{dt}$. Acceleration $a$ is the first derivative of velocity with respect to time $t$, given by $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$. First, find the time $t$ when acceleration $a = 0$, then substitute that time into the velocity expression.

For a particle moving in a straight line the position of the particle at time $(t)$ is given by $x = \frac{t^3}{6} - t^2 - 9t + 18\text{ m}$. What is the velocity of the particle when its acceleration is zero :

Concept: Velocity $v$ is defined as the rate of change of displacement $s$ with respect to time $t$, given by $v = \frac{ds}{dt}$.

A particle moves along a straight line such that at time $t$ its displacement from a fixed point $O$ on the line is $3t^2 - 2$. The velocity of the particle when $t = 2$ is:

Concept: The rate of change of temperature with respect to time is given by the derivative of temperature with respect to time, $\frac{dT}{dt}$.

Temperature of a body varies with time as $T = (T_0 + \alpha t^2 + \beta \sin t)\text{K}$, where $T_0$ is the temperature in Kelvin at $t = 0\text{ sec}$. $\&$ $\alpha = 2/\pi\text{ K/s}^2$ $\&$ $\beta = -4\text{ K}$, then rate of change of temperature at $t = \pi\text{ sec}$. is

Concept: When velocity $v$ is given as a function of position $x$, acceleration $a$ is obtained using the chain rule: $a = v \frac{dv}{dx}$.

The velocity of a particle moving on the x-axis is given by $v = x^2 + x$ where $v$ is in $\text{m/s}$ and $x$ is in $\text{m}$. Find its acceleration in $\text{m/s}^2$ when passing through the point $x = 2\text{m}$

Concept: Acceleration $a$ is defined as the rate of change of velocity with respect to time, $a = \frac{dv}{dt}$. By using the chain rule, acceleration can also be expressed in terms of displacement $x$ as $a = v \frac{dv}{dx}$.

The velocity of any particle is related with its displacement as $x = \sqrt{v+1}$. Calculate the acceleration of the particle at $x = 5\text{ m}$.

Concept: Velocity $v$ is the rate of change of displacement $v = \frac{dx}{dt}$. Acceleration $a$ is the derivative of velocity with respect to time $a = \frac{dv}{dt}$. By using calculus, displacement $x$ as a function of time $t$ can be found by integrating the differential equation $\frac{dx}{dt} = \alpha \sqrt{x}$, and then differentiating $x(t)$ twice with respect to time gives the acceleration.

The velocity of a particle moving in the positive direction of x-axis varies as $v = \alpha \sqrt{x}$ where $\alpha$ is a positive constant. Assuming that at $t = 0$, the particle was located at $x = 0$, find the acceleration of the particle as a function of time.

$\frac{1}{4} \alpha^2 t$

$\frac{1}{2} \alpha^2$

Concept: Velocity components are given by $v_x = \frac{dx}{dt}$ and $v_y = \frac{dy}{dt}$. Acceleration components are $a_x = \frac{d^2x}{dt^2}$ and $a_y = \frac{d^2y}{dt^2}$. The magnitude of total velocity is $v = \sqrt{v_x^2 + v_y^2}$.

A particle moves in the xy-plane with constant acceleration $a$ directed along the negative y-axis. The equation of trajectory of the particle is $y = px - qx^2$, where $p$ and $q$ are positive constants. Find the velocity of the particle at the origin of coordinates.

$\sqrt{\frac{a(p^2+1)}{2q}}$

$\sqrt{\frac{a(p^2-1)}{2q}}$

$\sqrt{\frac{ap^2}{2q}}$

$\sqrt{\frac{a}{2q(p^2+1)}}$

Concept: Average velocity $v_{\text{avg}}$ over a time interval from $t_1$ to $t_2$ is defined as total displacement divided by total time, given by the formula $v_{\text{avg}} = \frac{\int_{t_1}^{t_2} v \, dt}{\int_{t_1}^{t_2} dt}$.

A particle moves along a straight line path such that its magnitude of velocity is given by $v = (3t^2 - 6t)\text{ ms}^{-1}$, where $t$ is the time in seconds. If it is initially located at the origin $O$, then determine the magnitude of particle's average velocity in time interval from $t = 0$ to $t = 4\text{ s}$.

Concept: Average speed is defined as the total distance covered divided by the total time taken. Since velocity changes sign when the particle turns back, total distance is calculated by integrating the absolute value of velocity over the given time interval, $s = \int_{t_1}^{t_2} \vert{}v\vert{} \, dt$, and $\text{Average Speed} = \frac{\int_{t_1}^{t_2} \vert{}v\vert{} \, dt}{\int_{t_1}^{t_2} dt}$.

A particle moves along a straight line path such that its magnitude of velocity is given by $v = (3t^2 - 6t)\text{ ms}^{-1}$, where $t$ is the time in seconds. If it is initially located at the origin $O$, then determine the magnitude of particle's average speed in time interval from $t = 0$ to $t = 4\text{ s}$.

Concept: Position vector is given by $\vec{r} = x\hat{i} + y\hat{j}$ and velocity vector by $\vec{v} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}$. The angle $\theta$ between vectors $\vec{r}$ and $\vec{v}$ is calculated using the dot product formula $\cos\theta = \frac{\vec{r} \cdot \vec{v}}{\vert{}\vec{r}\vert{} \vert{}\vec{v}\vert{}}$.

The coordinates of a particle moving in a plane are given by $x = 3 \cos 2t$ and $y = 4 \sin 2t$. The angle between position vector $\vec{r}$ and velocity vector $\vec{v}$ at $t = \frac{\pi}{4}$ is:

Concept: To find the total distance travelled by a particle moving in a straight line, we must account for any changes in its direction of motion. The particle turns when its velocity $v = \frac{dx}{dt} = 0$. The total distance is the sum of the absolute magnitudes of displacement over each interval between turning points: $\text{Distance} = \vert{}x(t_1) - x(t_0)\vert{} + \vert{}x(t_2) - x(t_1)\vert{} + \dots + \vert{}x(t_n) - x(t_{n-1})\vert{}$.

A particle moves in a straight line according to the relation $x = \frac{t^3}{3} - \frac{5t^2}{2} + 6t$. Find the distance travelled by the particle up to $t = 4\text{ sec}$.

$\frac{16}{3}\text{ m}$

$\frac{17}{3}\text{ m}$

$\frac{22}{3}\text{ m}$