iCON Education - Motion in 1D DPT Test

Concept: Displacement is a vector quantity defined as the shortest distance between the initial position and the final position. When a body moves in three mutually perpendicular directions, the magnitude of the net displacement vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $r = \sqrt{x^2 + y^2 + z^2}$.

A body moves $6\text{ m}$ north, $8\text{ m}$ east and $10\text{ m}$ vertically upwards. What is its resultant displacement from initial position?

$10\sqrt{2}\text{ m}$

$\frac{10}{\sqrt{2}}\text{ m}$

$10 \times 2\text{ m}$

Concept: Displacement is a vector quantity representing the shortest straight-line distance from the initial position to the final position. When two displacement vectors are perpendicular to each other, the magnitude of the resultant displacement vector $\vec{r} = x\hat{i} + y\hat{j}$ is calculated using the Pythagorean theorem: $r = \sqrt{x^2 + y^2}$.

A man goes $10\text{ m}$ towards North, then $20\text{ m}$ towards east then displacement is

Concept: Displacement is a vector quantity representing the shortest path from the initial to the final position. A motion in a plane can be broken down into orthogonal vector components along the East-West (x-axis) and North-South (y-axis) directions. The total displacement is the vector sum of individual displacement vectors $\vec{r} = \vec{r}_1 + \vec{r}_2 + \vec{r}_3 = x\hat{i} + y\hat{j}$.

A person moves $30\text{ m}$ north and then $20\text{ m}$ towards east and finally $30\sqrt{2}\text{ m}$ in south-west direction. The displacement of the person from the origin will be

$10\text{ m}$ along north

$10\text{ m}$ along south

$10\text{ m}$ along west

Concept: Displacement is a vector quantity defined as the shortest distance from the initial position to the final position. In three-dimensional space, the net displacement vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is obtained by summing the individual vector displacements along mutually perpendicular axes. The magnitude of the displacement vector is given by $r = \sqrt{x^2 + y^2 + z^2}$.

An aeroplane flies $400\text{ m}$ north and $300\text{ m}$ south and then flies $1200\text{ m}$ upwards then net displacement is

Concept: Displacement is a vector quantity defined as the shortest straight-line distance from the initial position to the final position. For motion along a circular path, completing one full revolution brings the object back to its starting point, resulting in zero displacement for that complete turn. If an object stops at a diametrically opposite point after a fractional revolution, the magnitude of the displacement is equal to the diameter of the circular track ($2R$).

An athlete completes one round of a circular track of radius $R$ in $40\text{ sec}$. What will be his displacement at the end of $2\text{ min } 20\text{ sec}$?

Concept: When a circular wheel rolls forward without slipping, the point initially in contact with the ground undergoes both horizontal translation and vertical motion. During half a revolution, the horizontal displacement of the point is equal to half the circumference of the wheel ($\pi r$), and its vertical displacement from the bottom to the top position is equal to the diameter of the wheel ($2r$). The net displacement magnitude is calculated using the Pythagorean theorem: $S = \sqrt{x^2 + y^2}$.

A wheel of radius $1\text{ meter}$ rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is

Concept: Average velocity is defined as the total displacement divided by the total time taken. When a journey is divided into equal distance intervals covered at different uniform speeds, the average velocity is given by the harmonic mean of the speeds.

A person travels along a straight road for half the distance with velocity $v_1$ and the remaining half distance with velocity $v_2$. The average velocity is given by

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

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

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

Concept: Average speed is defined as the total distance covered divided by the total time taken. When a body travels equal distances at two different speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the two speeds: $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$.

A car travels from $A$ to $B$ at a speed of $20\text{ km/hr}$ and returns at a speed of $30\text{ km/hr}$. The average speed of the car for the whole journey is

Concept: Average speed is defined as the total distance covered divided by the total time taken for the journey. When a body covers equal distances for the forward and return journeys at different uniform speeds $v_1$ and $v_2$, the average speed is given by $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$.

A boy walks to his school at a distance of $6\text{ km}$ with constant speed of $2.5\text{ km/hr}$ and walks back with a constant speed of $4\text{ km/hr}$. His average speed for round trip expressed in $\text{km/hour}$ is

Concept: Average speed is defined as the total distance covered divided by the total time taken for the journey. When a journey is divided into two equal distances covered at different uniform speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the speeds: $v_{av} = \frac{2 v_1 v_2}{v_1 + v_2}$.

A car travels the first half of a distance between two places at a speed of $30\text{ km/hr}$ and the second half of the distance at $50\text{ km/hr}$. The average speed of the car for the whole journey is

Concept: Average speed is defined as the total distance covered divided by the total time taken for the trip. When a journey is divided into unequal distance segments traveled at different constant speeds, the total time is calculated by summing the time taken for each segment $t_i = \frac{s_i}{v_i}$. The average speed is then $v_{av} = \frac{S}{T_{total}}$.

One car moving on a straight road covers one third of the distance with $20\text{ km/hr}$ and the rest with $60\text{ km/hr}$. The average speed is

$46\frac{2}{3}\text{ km/hr}$

Concept: Average speed is defined as total distance divided by total time elapsed. When a body moves with uniform speed $v_1$ for time $t_1$ and uniform speed $v_2$ for time $t_2$, the average speed is given by the time-weighted average $v_{av} = \frac{v_1 t_1 + v_2 t_2}{t_1 + t_2}$. When the time intervals are equal ($t_1 = t_2 = \frac{t}{2}$), the average speed simplifies to the arithmetic mean of the speeds: $v_{av} = \frac{v_1 + v_2}{2}$.

A car moves for half of its time at $80\text{ km/h}$ and for rest half of time at $40\text{ km/h}$. Total distance covered is $60\text{ km}$. What is the average speed of the car?

Concept: Average speed is defined as the ratio of total distance traveled to total time taken. When a body moves with different speeds over different time intervals, the individual distances covered are calculated as $d = v \times t$. The average speed is then calculated using $v_{av} = \frac{d_1 + d_2}{t_1 + t_2}$.

A train has a speed of $60\text{ km/h}$ for the first one hour and $40\text{ km/h}$ for the next half hour. Its average speed in $\text{km/h}$ is

Concept: When a train crosses an object with a non-negligible length, such as a bridge, the total distance covered by the train to completely cross it is equal to the sum of the length of the train and the length of the bridge ($D = L_{train} + L_{bridge}$). The speed is converted from $\text{km/h}$ to $\text{m/s}$ using the conversion factor $1\text{ km/h} = \frac{5}{18}\text{ m/s}$, and the time required is determined using $t = \frac{D}{v}$.

A $150\text{ m}$ long train is moving with a uniform velocity of $45\text{ km/h}$. The time taken by the train to cross a bridge of length $850\text{ meters}$ is

Concept: Displacement is a vector quantity defined as the net change in position from the starting point to the ending point. If an object returns to its initial position, its total displacement is zero regardless of the distance traveled. Average speed is a scalar quantity calculated as total distance divided by total time: $v_{av} = \frac{\text{Total distance}}{\text{Total time}}$.

A particle is constrained to move on a straight line path. It returns to the starting point after $10\text{ sec}$. The total distance covered by the particle during this time is $30\text{ m}$. Which of the following statements about the motion of the particle is false?

Displacement of the particle is zero

Average speed of the particle is $3\text{ m/s}$

Displacement of the particle is $30\text{ m}$

Concept: Average speed is defined as the total distance covered divided by the total time taken ($v_{av} = \frac{\text{Total distance}}{\text{Total time}}$). When calculating average speed over a specific time interval, one must determine the position and distance traveled up to that exact time limit by considering the separate legs of the motion.

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

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

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

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

Concept: Average velocity is defined as total displacement divided by total time ($\vec{v}_{av} = \frac{\Delta\vec{r}}{\Delta t}$), while average speed is defined as total distance divided by total time ($v_{av} = \frac{\Delta s}{\Delta t}$). Since distance is always greater than or equal to the magnitude of displacement ($\text{Distance} \ge \vert{}\text{Displacement}\vert{}$), the magnitude of average velocity is always less than or equal to the average speed. Therefore, the ratio of average velocity to average speed is unity or less ($\le 1$).

The ratio of the numerical values of the average velocity and average speed of a body is always

Concept: Mean (or average) velocity is defined as total displacement divided by total time elapsed. When a body moves along a straight path with different uniform velocities for equal time intervals ($t_1 = t_2 = \frac{T}{2}$), the mean velocity is given by the arithmetic mean of the individual velocities: $V = \frac{v_1 + v_2}{2}$.

A person travels along a straight road for the first half time with a velocity $v_1$ and the next half time with a velocity $v_2$. The mean velocity $V$ of the man is

$\frac{2}{V} = \frac{1}{v_1} + \frac{1}{v_2}$

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

$V = \sqrt{v_1 v_2}$

$V = \sqrt{\frac{v_1}{v_2}}$

Concept: Average speed is defined as the total distance covered divided by the total time taken for the trip. When a journey is divided into distance fractions $f_1$ and $f_2$ covered at speeds $v_1$ and $v_2$ respectively, the total time is calculated by adding the time taken for each fraction: $T = \frac{f_1 S}{v_1} + \frac{f_2 S}{v_2}$. Average speed is then given by $v_{av} = \frac{S}{T}$.

If a car covers $2/5$ of the total distance with $v_1$ speed and $3/5$ distance with $v_2$ then average speed is

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

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

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

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

Concept: Average velocity is defined as the total displacement divided by the total time taken. When a body moves with velocities $v_1, v_2, \dots, v_n$ for equal intervals of time ($t_1 = t_2 = \dots = t_n = t$), the average velocity simplifies to the arithmetic mean of the individual velocities: $v_{av} = \frac{v_1 + v_2 + \dots + v_n}{n}$.

A particle moves for $20\text{ seconds}$ with velocity $3\text{ m/s}$ and then velocity $4\text{ m/s}$ for another $20\text{ seconds}$ and finally moves with velocity $5\text{ m/s}$ for next $20\text{ seconds}$. What is the average velocity of the particle?

Concept: When an object covers two equal distances with different uniform velocities $v_1$ and $v_2$, the average velocity over the total distance is given by the harmonic mean of the two velocities:
$v_{avg} = \frac{2 v_1 v_2}{v_1 + v_2}$

A car travels half the distance with constant velocity of 40 km/h and the remaining half with a constant velocity of 60 km/h. The average velocity of the car in km/h is: