iCON Education - Kinematics-1D Error less Test

Concept: Displacement is a vector quantity defined as the shortest straight-line distance from the initial position to the final position. If East is taken along the positive x-axis and North along the positive y-axis, the displacement vector is given by $\vec{r} = x\hat{i} + y\hat{j}$. Its magnitude is calculated using the formula $\vert{}\vec{r}\vert{} = \sqrt{x^2 + y^2}$.

A man goes 10 m towards North, then 20 m towards East. What is his displacement?

Concept: Distance is the actual total path length covered during motion, which for a quarter circle of radius $r$ is one fourth of the circumference, given by $s = \frac{2\pi r}{4} = \frac{\pi r}{2}$.
Displacement is the straight-line distance from the initial position to the final position. In a quarter circular path, the initial and final position vectors form a right-angled triangle with the center, so the magnitude of displacement is $d = \sqrt{r^2 + r^2} = r\sqrt{2}$.

A body moves over one fourth of a circular arc in a circle of radius $r$. The magnitude of distance travelled and displacement will be respectively

$\frac{\pi r}{2}, r\sqrt{2}$

$\frac{\pi r}{4}, r$

$\pi r, \frac{r}{\sqrt{2}}$

Concept: When a wheel rolls forward through half a revolution without slipping:
1. The horizontal displacement of the center of the wheel (and thus the whole wheel) is equal to half of its circumference, which is $x = \pi R$.
2. The point initially in contact with the ground moves to the top of the wheel after half a revolution, causing a vertical displacement equal to the diameter of the wheel, $y = 2R$.
The net displacement vector connects the initial ground contact point to its final top position. By the Pythagorean theorem, the magnitude of total displacement is $d = \sqrt{x^2 + y^2} = \sqrt{(\pi R)^2 + (2R)^2} = R\sqrt{\pi^2 + 4}$.

The displacement of the point of the wheel initially in contact with the ground, when the wheel rolls forward half a revolution will be (radius of the wheel is $R$)

$\frac{R}{\sqrt{\pi^2 + 4}}$

Concept: Average speed is defined as the total distance travelled divided by the total time taken. When a journey is divided into segments, the time for each segment is calculated using the formula $t = \frac{\text{distance}}{\text{speed}}$, and the overall average speed is given by $v_{\text{av}} = \frac{d_{\text{total}}}{t_{\text{total}}}$.

If a car covers $\frac{2}{5}\text{th}$ of the total distance with $v_1$ speed and $\frac{3}{5}\text{th}$ distance with $v_2$, then average speed is

$\frac{1}{2}\sqrt{v_1 v_2}$

$\frac{v_1 + v_2}{2}$

$\frac{2v_1 v_2}{v_1 + v_2}$

$\frac{5v_1 v_2}{3v_1 + 2v_2}$

Concept: Displacement is defined as the shortest distance between the initial position and the final position of an object. If a body starts its motion from a point and returns to the exact same point, its net displacement is zero, which makes its average velocity zero since $\text{Average Velocity} = \frac{\text{Net Displacement}}{\text{Total Time}}$. However, the distance covered along the path is non-zero, meaning the speed increases.

A car accelerated from initial position and then returned at initial point, then

Velocity is zero but speed increases

Speed is zero but velocity increases

Both speed and velocity increase

Both speed and velocity decrease

Concept: Average speed is calculated as the total distance covered divided by the total time elapsed: $v_{\text{av}} = \frac{\text{Total Distance}}{\text{Total Time}}$. To find the distance travelled in a specific time interval, determine the distance covered in each leg of the journey using $d = v \times t$.

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. The average speed of the man over the interval of time 0 to 40 min is equal to

$\frac{25}{4} \text{ km/h}$

$\frac{30}{4} \text{ km/h}$

$\frac{45}{8} \text{ km/h}$

Concept: Average velocity over a time interval $\Delta t$ is defined as $v_{\text{av}} = \frac{\Delta x}{\Delta t}$. If the velocity increases by equal amounts in equal time intervals, the acceleration is constant, which signifies uniform accelerated motion.

The position of a particle moving along the x-axis at certain times is given below:

Which of the following describes the motion correctly?

Uniform, accelerated

Uniform, decelerated

Non-uniform, accelerated

There is not enough data for generalisation

Concept: In uniform motion, an object covers equal distances in equal intervals of time, meaning its speed (or velocity) remains constant. Since speed is given by the slope of the distance-time (or displacement-time) graph ($v = \frac{ds}{dt}$), uniform motion is represented by a straight line with a constant positive slope in a distance-time graph.

Which of the following graphs represents uniform motion?

Concept: In a displacement-time ($s-t$) graph, the slope of the line represents the velocity of the particle. The slope of a line inclined at an angle $\theta$ with the time axis is given by $\tan\theta$. Therefore, the velocity $v$ is equal to $\tan\theta$.

The displacement-time graph for two particles $A$ and $B$ are straight lines inclined at angles of $30^\circ$ and $60^\circ$ with the time axis. The ratio of velocities of $v_A : v_B$ is

Concept: In kinematics, velocity is the rate of change of displacement with respect to time, $v = \frac{dx}{dt}$, which is represented by the slope of the displacement-time graph with respect to the time axis. Therefore, $v = \tan\theta$, where $\theta$ is the angle made by the line with the time axis.

From the following displacement-time graph, find out the velocity of a moving body:
(A graph is given with time on the vertical axis and displacement on the horizontal axis, showing a straight line making an angle of $30^\circ$ with the displacement axis).

$\frac{1}{\sqrt{3}} \text{ m/s}$

$\sqrt{3} \text{ m/s}$

$\frac{1}{3} \text{ m/s}$

Concept: Average velocity over a time interval is defined as the total displacement divided by the total time taken: $v_{\text{av}} = \frac{\Delta x}{\Delta t} = \frac{x(t_f) - x(t_i)}{t_f - t_i}$, where $x(t_f)$ is the final position at time $t_f$ and $x(t_i)$ is the initial position at time $t_i$.

The diagram shows the displacement-time graph for a particle moving in a straight line. The average velocity for the interval $t = 0$ to $t = 5$ is

$-2 \text{ ms}^{-1}$

Concept: In a displacement-time graph, the speed of a moving body is represented by the magnitude of the slope of the graph, given by $v = \left\vert{} \frac{\Delta y}{\Delta x} \right\vert{} = \left\vert{} \frac{\text{Change in displacement}}{\text{Change in time}} \right\vert{}$.

Figure shows the displacement-time graph of a body. What is the ratio of the speed in the first second and that in the next two seconds?

Concept: When air resistance is not ignored, the air drag force always opposes the direction of motion. During upward motion, both gravity and air resistance act downwards, giving a higher magnitude of acceleration $(g + a)$. During downward motion, gravity acts downwards while air resistance acts upwards, resulting in a lower magnitude of acceleration $(g - a)$.

A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the ball during its flight if the air resistance is not ignored?

Concept: In a speed-time graph, the acceleration at any segment is given by the slope of the line representing that segment, $a = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}$. The maximum acceleration corresponds to the segment with the steepest positive slope.

A train moves from one station to another in 2 hours time. Its speed-time graph during this motion is shown in the figure. The maximum acceleration during the journey is

$100 \text{ km h}^{-2}$

$160 \text{ km h}^{-2}$

$140 \text{ km h}^{-2}$

$120 \text{ km h}^{-2}$

Concept: The slope of a displacement-time ($s-t$) graph represents the instantaneous velocity ($v = \frac{ds}{dt}$). For a downward-opening parabolic $s-t$ graph, the slope decreases continuously at a constant rate, starting from a positive value, reaching zero at the peak, and then becoming increasingly negative. This indicates a uniform negative acceleration, represented by a straight line with a negative slope on a velocity-time graph.

The graph of displacement $v/s$ time shows a downward-opening parabolic curve starting from the origin. Its corresponding velocity-time graph will be: