iCON Education - Kinematics-1D

Q1 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Distance is the total length of the path covered by a body, regardless of direction. Displacement is the shortest straight-line distance between the initial and final positions, directed from the initial point to the final point. For a straight path, distance equals the magnitude of displacement. For a semicircular path of radius $r$, the path distance is $\pi r$, while the magnitude of displacement is the diameter $2r$.

Ram takes path 1 (straight line) to go from $P$ to $Q$ and Shyam takes path 2 (semicircle) of diameter $100\text{ m}$.

(a) Find the distance travelled by Ram and Shyam.
(b) Find the displacement of Ram and Shyam.

A.

(a) Ram $= 100\text{ m}$, Shyam $= 100\text{ m}$; (b) Ram $= 50\pi\text{ m}$, Shyam $= 100\text{ m}$

B.

(a) Ram $= 100\text{ m}$, Shyam $= 50\pi\text{ m}$; (b) Ram $= 100\text{ m}$, Shyam $= 100\text{ m}$

C.

(a) Ram $= 50\pi\text{ m}$, Shyam $= 100\text{ m}$; (b) Ram $= 100\text{ m}$, Shyam $= 50\pi\text{ m}$

D.

(a) Ram $= 100\text{ m}$, Shyam $= 100\pi\text{ m}$; (b) Ram $= 50\text{ m}$, Shyam $= 50\text{ m}$

Q2 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Displacement is defined as the change in position of a particle ($\Delta x = x_{\text{final}} - x_{\text{initial}}$). Distance is the actual total length of the path traversed by the body. To find the position at any time $t$, express $x$ in terms of $t$ as $x = (t - 3)^2$.

The position $x$ (in metre) of a particle varies with time $t$ (in second) as $t = \sqrt{x} + 3$. Calculate the:
(a) displacement of the particle from $t = 0$ to $t = 3\text{ s}$
(b) displacement of the particle from $t = 3$ to $t = 6\text{ s}$
(c) distance travelled and displacement from $t = 0$ to $t = 6\text{ s}$

A.

(a) $-9\text{ m}$, (b) $+9\text{ m}$, (c) distance $= 18\text{ m}$, displacement $= 0\text{ m}$

B.

(a) $+9\text{ m}$, (b) $-9\text{ m}$, (c) distance $= 9\text{ m}$, displacement $= 18\text{ m}$

C.

(a) $-9\text{ m}$, (b) $+9\text{ m}$, (c) distance $= 0\text{ m}$, displacement $= 18\text{ m}$

D.

(a) $0\text{ m}$, (b) $+9\text{ m}$, (c) distance $= 18\text{ m}$, displacement $= 9\text{ m}$

Q3 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Position Vector and Displacement Vector in 3D Coordinate System.
If a particle is located at point $(x, y, z)$, its position vector is $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. The displacement vector between initial position $\vec{r}_1$ and final position $\vec{r}_2$ is $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1 = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.

A particle starts from point $(1, 1, 1)\text{ m}$ and reaches at point $(4, 5, 13)\text{ m}$. Find the initial position vector, final position vector, and displacement vector.

A.

Initial: $4\hat{i} + 5\hat{j} + 13\hat{k}$, Final: $\hat{i} + \hat{j} + \hat{k}$, Displacement: $-3\hat{i} - 4\hat{j} - 12\hat{k}$

B.

Initial: $\hat{i} + \hat{j} + \hat{k}$, Final: $4\hat{i} + 5\hat{j} + 13\hat{k}$, Displacement: $5\hat{i} + 6\hat{j} + 14\hat{k}$

C.

Initial: $\hat{i} + \hat{j} + \hat{k}$, Final: $4\hat{i} + 5\hat{j} + 13\hat{k}$, Displacement: $3\hat{i} + 4\hat{j} + 12\hat{k}$

D.

Initial: $3\hat{i} + 4\hat{j} + 12\hat{k}$, Final: $4\hat{i} + 5\hat{j} + 13\hat{k}$, Displacement: $\hat{i} + \hat{j} + \hat{k}$

Q4 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Displacement is a vector quantity representing the shortest distance between the initial and final positions. In a three-dimensional Cartesian coordinate system, if a displacement vector is represented as $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, its magnitude is given by $\vert{}\vec{r}\vert{} = \sqrt{x^2 + y^2 + z^2}$.

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?

A.

$10\text{ m}$

B.

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

C.

$20\text{ m}$

D.

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

Q5 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Displacement is the shortest straight-line distance from the initial position to the final position. Since North and East directions are perpendicular ($90^\circ$ to each other), the magnitude of displacement can be calculated using the Pythagorean theorem: $s = \sqrt{x^2 + y^2}$, where $x$ and $y$ are perpendicular components of displacement.

A man goes $10\text{ m}$ towards North, then $20\text{ m}$ towards East. What is his displacement?

A.

$25.5\text{ m}$

B.

$22.36\text{ m}$

C.

$30.0\text{ m}$

D.

$25.0\text{ m}$

Q6 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: The displacement vector between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in a three-dimensional coordinate system is given by $\vec{r} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.

An aeroplane flies from $P(-4\text{ m}, -5\text{ m}, +8\text{ m})$ to $Q(7\text{ m}, -2\text{ m}, -3\text{ m})$ in an xyz coordinate system. The displacement position vector of the aeroplane is

A.

$11\hat{i} + 3\hat{j} + 11\hat{k}$

B.

$11\hat{i} - 3\hat{j} + 11\hat{k}$

C.

$11\hat{i} + 3\hat{j} - 11\hat{k}$

D.

$11\hat{i} - 3\hat{j} - 11\hat{k}$

Q7 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Displacement is the shortest straight-line distance between the initial position ($A$) and final position ($B$). In a quarter-circle path, it forms a right-angled triangle with the radius vectors as legs, so the magnitude of displacement $= \sqrt{R^2 + R^2} = \sqrt{2}R$.
Distance is the total path length traveled along the arc of the circle. The arc length for three-quarters of a full circle ($270^\circ$) from $A$ clockwise to $B$ is $\frac{3}{4} \times 2\pi R = \frac{3\pi R}{2}$.

A body moves in circular path of radius $R$ from $A$ to $B$ as shown. Its displacement and distance covered are

A.

$R, \frac{3\pi R}{2}$

B.

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

C.

$\sqrt{2}R, \frac{3\pi R}{2}$

D.

None of these

Q8 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Distance is the total path length traveled along the semicircular arc, calculated as $d = \pi r$. Displacement is the straight-line distance between the initial and final position, which equals the diameter of the circle, $s = 2r$.

A particle covers half of the circle of radius $r$. Then the displacement and distance of the particle are respectively

A.

$2\pi r, 0$

B.

$2r, \pi r$

C.

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

D.

$\pi r, r$

Q9 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Distance is the total path length traveled by a particle, given by the scalar sum of individual path lengths.
Displacement is the shortest distance between the initial and final positions, represented as a vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ with magnitude $\vert{}\vec{r}\vert{} = \sqrt{x^2 + y^2 + z^2}$.

A particle moves $5\text{ m}$ along east, $6\text{ m}$ along north, and $10\text{ m}$ in the upward direction. Find its distance and displacement.

A.

Distance = $21\text{ m}$, Displacement = $15\text{ m}$

B.

Distance = $15\text{ m}$, Displacement = $\sqrt{161}\text{ m}$

C.

Distance = $21\text{ m}$, Displacement = $\sqrt{161}\text{ m}$

D.

Distance = $21\text{ m}$, Displacement = $21\text{ m}$

Q10 DPT Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Distance is the scalar sum of all individual path lengths covered during motion.
Displacement is the vector sum of individual displacements, representing the net change in position: $\vec{d} = x\hat{i} + y\hat{j}$.

A particle moves $10\text{ m}$ east, $5\text{ m}$ north, $6\text{ m}$ south, $8\text{ m}$ west, $15\text{ m}$ east, and $20\text{ m}$ north. Find its total distance and displacement vector.

A.

Distance = $64\text{ m}$, Displacement = $17\hat{i} + 19\hat{j}\text{ m}$

B.

Distance = $58\text{ m}$, Displacement = $17\hat{i} + 19\hat{j}\text{ m}$

C.

Distance = $64\text{ m}$, Displacement = $19\hat{i} + 17\hat{j}\text{ m}$

D.

Distance = $60\text{ m}$, Displacement = $10\hat{i} + 20\hat{j}\text{ m}$

Q11 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Average Speed for Equal Distance Intervals
$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a particle covers two equal distances $S$ with speeds $v_1$ and $v_2$, the total time taken is the sum of the times taken for each half ($t_1 = \frac{S}{v_1}$ and $t_2 = \frac{S}{v_2}$). The average speed is given by the harmonic mean of the two speeds.

A particle travels half of total distance with speed $v_1$ and next half with speed $v_2$ along a straight line. Find out the average speed of the particle.

A.

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

B.

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

C.

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

D.

$\sqrt{v_1 v_2}$

Q12 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: When a body covers two equal distances with speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

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

A.

$42.5\text{ km/hr}$

B.

$40.0\text{ km/hr}$

C.

$37.5\text{ km/hr}$

D.

$35.0\text{ km/hr}$

Q13 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: When a body travels a certain distance $d$ at speed $v_1$ and returns over the same distance $d$ at speed $v_2$, the total distance is $2d$. The average speed is given by the harmonic mean of the two speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

A car travels from $A$ to $B$ at a speed of $20\text{ km h}^{-1}$ and returns at a speed of $30\text{ km h}^{-1}$. The average speed of the car for the whole journey is

A.

$5\text{ km h}^{-1}$

B.

$24\text{ km h}^{-1}$

C.

$25\text{ km h}^{-1}$

D.

$50\text{ km h}^{-1}$

Q14 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Average Speed for Equal Distances
When a body covers equal distances with different speeds $v_1$ and $v_2$, the average speed for the round trip is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

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

A.

$25\text{ km/hr}$

B.

$24\text{ km/hr}$

C.

$50\text{ km/hr}$

D.

$5\text{ km/hr}$

Q15 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Average Speed for Equal Distances
When a body covers equal distances with different speeds $v_1$ and $v_2$, the average speed for the total journey is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

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

A.

$\frac{24}{13}$

B.

$\frac{40}{13}$

C.

$3$

D.

$\frac{1}{2}$

Q16 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: When a body covers two equal distance intervals with speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

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

A.

$42.5\text{ km/hr}$

B.

$40.0\text{ km/hr}$

C.

$37.5\text{ km/hr}$

D.

$35.0\text{ km/hr}$

Q17 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Average Speed in a Given Time Interval
$$\text{Average Speed} = \frac{\text{Total Distance Covered}}{\text{Total Time Taken}}$$
To find the average speed over a given time interval, calculate the total distance covered during that exact time period and divide it by the duration of the time interval.

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

A.

$5\text{ km/h}$

B.

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

C.

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

D.

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

Q18 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Average Speed with Variable Distances and Speeds
$$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a path is divided into different distance fractions with different constant speeds, calculate the time taken for each section and divide the total distance by the sum of individual time intervals.

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

A.

$40\text{ km/hr}$

B.

$80\text{ km/hr}$

C.

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

D.

$36\text{ km/hr}$

Q19 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Average speed is defined as the ratio of total distance traveled to total time taken:
$$v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a journey is divided into segments of known distances and speeds, the total time is calculated by summing the time taken for each individual segment.

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

A.

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

B.

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

C.

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

D.

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

Q20 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: Average Velocity for Equal Time Intervals
$$\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$$
When an object moves with different uniform velocities over equal consecutive time intervals $t$, the average velocity is equal to the arithmetic mean of the individual velocities.

A person travelling on a straight line moves with a uniform velocity $v_1$ for some time and with uniform velocity $v_2$ for the next equal time. The average velocity $v$ is given by

A.

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

B.

$v = \sqrt{v_1 v_2}$

C.

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

D.

$v = \frac{v_1 v_2}{v_1 + v_2}$

Q21 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
Average speed is a scalar quantity defined as the total path length (distance) divided by the total time taken. Unlike velocity, direction does not affect distance; the distance covered in each segment is simply speed multiplied by time.

A man walks for some time $t$ with velocity $v$ due east. Then he walks for same time $t$ with velocity $v$ due north. The average speed of the man is

A.

$2v$

B.

$\sqrt{2}v$

C.

$v$

D.

$\frac{v}{\sqrt{2}}$

Q22 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
$$\text{Average Velocity} = \frac{\text{Net Displacement}}{\text{Total Time}}$$
Average speed depends on the total path length traveled, whereas average velocity depends on the net displacement between the initial and final positions. When an object returns to its starting point, its net displacement is zero.

A car travels a distance $d$ on a straight road in two hours and then returns to the starting point in next three hours. Its average speed and average velocity is

A.

$\frac{d}{5}, 0$

B.

$\frac{2d}{5}, 0$

C.

$\frac{5d}{6}, \frac{d}{5}$

D.

none of these

Q23 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Velocity} = \frac{\text{Net Displacement}}{\text{Total Time}}$$
Net displacement is the shortest straight-line distance from the initial position to the final position. For perpendicular displacements along two axes ($x$ and $y$), the total displacement magnitude is $\sqrt{x^2 + y^2}$.

A particle moves in the east direction with $15\text{ m/s}$ for $2\text{ s}$ then northwards with $5\text{ m/s}$ for $8\text{ s}$. Average velocity of the particle is

A.

$1\text{ m/s}$

B.

$5\text{ m/s}$

C.

$7\text{ m/s}$

D.

$10\text{ m/s}$

Q24 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a body covers equal distances with speeds $v_1$ and $v_2$, the average speed is given by the harmonic mean of the speeds: $v_{\text{avg}} = \frac{2v_1 v_2}{v_1 + v_2}$.

An object travels $10\text{ km}$ at a speed of $100\text{ m/s}$ and another $10\text{ km}$ at $50\text{ m/s}$. The average speed over the whole distance is

A.

$75\text{ m/s}$

B.

$55\text{ m/s}$

C.

$66.7\text{ m/s}$

D.

$33.3\text{ m/s}$

Q25 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$$
When a particle moves in equal time intervals $t$ with different velocities $v_1, v_2, v_3, \dots, v_n$ along a straight line in the same direction, the average velocity is equal to the arithmetic mean of the individual velocities:
$$v_{\text{avg}} = \frac{v_1 + v_2 + \dots + v_n}{n}$$

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

A.

$3\text{ m/s}$

B.

$4\text{ m/s}$

C.

$5\text{ m/s}$

D.

$\text{Zero}$

Q26 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a total distance $S$ is divided into equal distance parts traveled at different speeds $v_1, v_2, v_3, \dots, v_n$, the average speed is given by the harmonic mean of the individual speeds:
$$v_{\text{avg}} = \frac{n}{\frac{1}{v_1} + \frac{1}{v_2} + \dots + \frac{1}{v_n}}$$

A body has speed $V$, $2V$ and $3V$ in first $1/3$ part of total travelled distance $S$, second $1/3$ part of $S$ and third $1/3$ part of $S$ respectively. Its average speed will be

A.

$V$

B.

$2V$

C.

$\frac{18}{11}V$

D.

$\frac{11}{18}V$

Q27 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$$
When time is divided into equal intervals $t$, and an object moves with uniform velocities $v_1, v_2, \dots, v_n$ during each interval, the average velocity is given by the arithmetic mean of the individual velocities:
$$v_{\text{avg}} = \frac{v_1 + v_2 + \dots + v_n}{n}$$

A body covers one-third of the time with a velocity $v_1$, the second one-third of the time with a velocity $v_2$, and the last one-third of the time with a velocity $v_3$. The average velocity is

A.

$\frac{v_1 + v_2 + v_3}{3}$

B.

$\frac{3v_1 v_2 v_3}{v_1 v_2 + v_2 v_3 + v_3 v_1}$

C.

$\frac{v_1 v_2 + v_2 v_3 + v_3 v_1}{3}$

D.

$\frac{v_1 v_2 v_3}{3}$

Q28 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a journey consists of different sections, first calculate the effective average speed for any sub-segment involving equal time intervals:
$$v_{\text{eq time}} = \frac{v_1 + v_2}{2}$$
Then use the formula for equal distance segments to find the total average speed:
$$v_{\text{avg}} = \frac{2 v_{\text{first half}} v_{\text{eq time}}}{v_{\text{first half}} + v_{\text{eq time}}}$$

A particle moving in a straight line covers half the distance with speed of $10\text{ m/s}$. The other half of the distance is covered in two equal time intervals with speed of $4.5\text{ m/s}$ and $7.5\text{ m/s}$ respectively. The average speed of the particle during this motion is

A.

$8.0\text{ m/s}$

B.

$12.0\text{ m/s}$

C.

$10.0\text{ m/s}$

D.

$7.5\text{ m/s}$

Q29 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$$
For a multi-stage motion:
1. For equal time intervals, the effective velocity over that part is the arithmetic mean: $v_{\text{eff}} = \frac{v_1 + v_2}{2}$.
2. For two equal distance parts covered with speeds $v_0$ and $v_{\text{eff}}$, the average velocity is given by their harmonic mean: $v_{\text{avg}} = \frac{2 v_0 v_{\text{eff}}}{v_0 + v_{\text{eff}}}$.

A point object traverses half the distance with velocity $v_0$. The remaining part of the distance is covered with velocity $v_1$ for half the time and with velocity $v_2$ for the rest half. The average velocity of the object for the whole journey is

A.

$\frac{2v_1 (v_0 + v_2)}{v_0 + 2v_1 + 2v_2}$

B.

$\frac{2v (v_0 + v_1)}{v_0 + v_1 + v_2}$

C.

$\frac{2v_0 (v_1 + v_2)}{v_1 + v_2 + 2v_0}$

D.

$\frac{2v_2 (v_0 + v_1)}{v_1 + 2v_2 + v_0}$

Q30 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 21 Jul 2026

Concept: $$\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$$
When a total distance is divided into parts with different constant speeds, the average speed is found by taking the total distance divided by the sum of the time intervals taken for each part.

If a train travels $\frac{1}{4}\text{th}$ of the total distance with speed $v_1$ and the remaining $\frac{3}{4}\text{th}$ distance with speed $v_2$, then its average speed is

A.

$\frac{4v_1 v_2}{3v_1 + v_2}$

B.

$\frac{v_1 + 3v_2}{4}$

C.

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

D.

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

Q31 Resnick Haliday v-t Graph MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Jul 2026

Concept: The change in position (displacement $\Delta x$) of a particle moving in one dimension over a time interval $t_1$ to $t_2$ is equal to the area under the velocity-time graph $v_x(t)$ between $t_1$ and $t_2$:
$\Delta x = x(t_2) - x(t_1) = \int_{t_1}^{t_2} v_x(t) \, dt$

$v_x$ is the velocity of a particle moving along the $x$-axis as shown in the figure. If $x = 2.0\text{ m}$ at $t = 1.0\text{ s}$, what is the position of the particle at $t = 6.0\text{ s}$?

A.

$-2.0\text{ m}$

B.

$+2.0\text{ m}$

C.

$+1.0\text{ m}$

D.

$-1.0\text{ m}$

Q32 Resnick Haliday Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 23 Jul 2026

Concept: Average acceleration $a_{\text{avg}}$ is defined as the change in velocity $\Delta v$ divided by the total time interval $\Delta t$:
$a_{\text{avg}} = \frac{v_f - v_i}{\Delta t}$
where $v_i$ is the initial velocity and $v_f$ is the final velocity.

What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of $8.0\text{ m/s}$ when going down a slope for $5.0\text{ s}$?

A.

$1.1\text{ m/s}^2$

B.

$1.9\text{ m/s}^2$

C.

$1.6\text{ m/s}^2$

D.

$0.85\text{ m/s}^2$

Q33 Error less Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

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?

A.

22.5 m

B.

25 m

C.

25.5 m

D.

30 m

Q34 Error less Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

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

A.

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

B.

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

C.

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

D.

$\pi r, r$

Q35 Error less Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

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 $R$)

A.

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

B.

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

C.

$2\pi R$

D.

$\pi R$

Q36 Error less Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

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

A.

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

B.

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

C.

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

D.

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

Q37 Error less Distance & Displacement MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

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

A.

Velocity is zero but speed increases

B.

Speed is zero but velocity increases

C.

Both speed and velocity increase

D.

Both speed and velocity decrease

Q38 Error less Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

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

A.

5 km/h

B.

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

C.

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

D.

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

Q39 DPT Velocity & Acceleration MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Instantaneous velocity is the rate of change of position with respect to time, given by the derivative of position $x$ with respect to time $t$, $v = \frac{dx}{dt}$. To find the velocity at a specific moment, differentiate the position function and substitute the given value of time $t$.

The motion of a particle is described by the equation $x = a + bt^2$ where $a = 15 \text{ cm}$ and $b = 3 \text{ cm/s}^2$. Its instantaneous velocity at time $3 \text{ sec}$ will be

A.

36 cm/sec

B.

18 cm/sec

C.

16 cm/sec

D.

32 cm/sec

Q40 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Average speed is calculated by dividing the total distance travelled by the total time taken for the journey: $v_{\text{av}} = \frac{\text{Total Distance}}{\text{Total Time}}$. The distance for each interval is calculated using the formula $d = v \times t$.

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

A.

50

B.

53.33

C.

48

D.

70

Q41 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: When a total distance is divided into equal distance parts, the average speed is given by the harmonic mean of the individual speeds. Total average speed is calculated using the formula $v_{\text{av}} = \frac{\text{Total Distance}}{\text{Total Time}}$.

A person completes half of his journey with speed $v_1$ and rest half with speed $v_2$. The average speed of the person is

A.

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

B.

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

C.

$v = \frac{v_1 v_2}{v_1 + v_2}$

D.

$v = \sqrt{v_1 v_2}$

Q42 DPT Avg Speed and Velocity MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Average speed is defined as the total distance covered divided by the total time taken for the journey: $v_{\text{av}} = \frac{\text{Total Distance}}{\text{Total Time}}$. Time for each segment of the journey is calculated using the relation $t = \frac{\text{distance}}{\text{speed}}$.

A car moving on a straight road covers one third of the distance with 20 km/hr and the rest with 60 km/hr. The average speed is

A.

40 km/hr

B.

80 km/hr

C.

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

D.

36 km/hr

Q43 DPT Velocity & Acceleration MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Acceleration is defined as the second derivative of displacement with respect to time, $a = \frac{d^2 s}{dt^2}$, or the rate of change of velocity with respect to time, $a = \frac{dv}{dt}$ where $v = \frac{ds}{dt}$.

The displacement of a particle, moving in a straight line, is given by $s = 2t^2 + 2t + 4$ where $s$ is in metres and $t$ in seconds. The acceleration of the particle is

A.

$2 \text{ m/s}^2$

B.

$4 \text{ m/s}^2$

C.

$6 \text{ m/s}^2$

D.

$8 \text{ m/s}^2$

Q44 DPT Velocity & Acceleration MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Acceleration is defined as the second time derivative of position, $a = \frac{d^2 x}{dt^2}$. To find the time when acceleration becomes zero, differentiate the position function $x(t)$ twice with respect to time $t$ to obtain the acceleration function, set $a(t) = 0$, and solve for $t$.

The position $x$ of a particle varies with time $t$ as $x = at^2 - bt^3$. The acceleration of the particle will be zero at time $t$ equal to

A.

$\frac{a}{b}$

B.

$\frac{2a}{3b}$

C.

$\frac{a}{3b}$

D.

Zero

Q45 DPT Velocity & Acceleration MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Initial velocity is the velocity of the particle at time $t = 0$, obtained by taking the first derivative of displacement with respect to time, $v = \frac{dy}{dt}$. Initial acceleration is the acceleration at $t = 0$, obtained by taking the second derivative of displacement with respect to time, $a = \frac{d^2 y}{dt^2} = \frac{dv}{dt}$.

The displacement of the particle is given by $y = a + bt + ct^2 - dt^4$. The initial velocity and acceleration are respectively

A.

$b, -4d$

B.

$-b, 2c$

C.

$b, 2c$

D.

$2c, -4d$

Q46 DPT Velocity & Acceleration MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Retardation is the magnitude of negative acceleration, $a = -\text{retardation}$. Velocity is the rate of change of position with respect to time, $v = \frac{dx}{dt}$, which can also be written as $v = \frac{1}{\frac{dt}{dx}}$. Acceleration can be expressed in terms of velocity and position using the chain rule: $a = \frac{dv}{dt} = v \frac{dv}{dx}$.

The relation between time $t$ and distance $x$ is $t = \alpha x^2 + \beta x$, where $\alpha$ and $\beta$ are constants. The retardation is ($v$ is the velocity)

A.

$2\alpha v^3$

B.

$2\beta v^3$

C.

$2\alpha\beta v^3$

D.

$2\beta^2 v^3$

Q47 DPT Velocity & Acceleration MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: When the displacement $x$ of a particle is directly proportional to the square of time $t^2$ (i.e., $x \propto t^2$), we can write $x = Kt^2$ where $K$ is a constant. The velocity is the first time derivative of displacement, $v = \frac{dx}{dt}$, and acceleration is the second time derivative of displacement, $a = \frac{d^2x}{dt^2}$. If acceleration is constant and independent of time, the motion is described as having uniform acceleration.

If displacement of a particle is directly proportional to the square of time, then the particle is moving with

A.

Variable acceleration

B.

Uniform acceleration

C.

Uniform velocity

D.

Variable acceleration but uniform velocity

Q48 DPT Velocity & Acceleration MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Average acceleration is defined as the change in velocity divided by the time taken, $\vec{a}_{\text{av}} = \frac{\Delta\vec{v}}{\Delta t} = \frac{\vec{v}_2 - \vec{v}_1}{\Delta t}$. Since velocity is a vector quantity, vector subtraction must be used to find $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$. The direction of the average acceleration is along the direction of the change in velocity vector $\Delta\vec{v}$.

A particle is moving eastwards with velocity of $5 \text{ m/s}$. In $10 \text{ sec}$ the velocity changes to $5 \text{ m/s}$ northwards. The average acceleration in this time is

A.

Zero

B.

$\frac{1}{\sqrt{2}} \text{ m/s}^2$ toward north-west

C.

$\frac{1}{\sqrt{2}} \text{ m/s}^2$ toward north-east

D.

$\frac{1}{2} \text{ m/s}^2$ toward north-west

Q49 DPT Velocity & Acceleration MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Acceleration is defined as the rate of change of velocity with respect to time, $a = \frac{\Delta v}{t} = \frac{v_2 - v_1}{t}$. Velocity is a vector quantity, so direction must be taken into account when defining initial and final velocities.

A body of mass 10 kg is moving with a constant velocity of $10 \text{ m/s}$. When a constant force acts for $4 \text{ sec}$ on it, it moves with a velocity $2 \text{ m/sec}$ in the opposite direction. The acceleration produced in it is

A.

$3 \text{ m/s}^2$

B.

$-3 \text{ m/s}^2$

C.

$0.3 \text{ m/s}^2$

D.

$-0.3 \text{ m/s}^2$

Q50 DPT Velocity & Acceleration MCQ

iCON Education HYD, 79930 92826, 73309 72826 25 Jul 2026

Concept: Displacement is obtained by integrating velocity with respect to time, $x = \int v \, dt$. To find the acceleration, we differentiate velocity with respect to time, $a = \frac{dv}{dt}$. When given the displacement, we first solve for the time $t$ using the position function and then substitute that value of $t$ into the acceleration expression.

A body starts from the origin and moves along the x-axis such that velocity at any instant is given by $(4t^3 - 2t)$, where $t$ is in second and velocity is in m/s. What is the acceleration of the particle, when it is $2 \text{ m}$ from the origin?

A.

$28 \text{ m/s}^2$

B.

$22 \text{ m/s}^2$

C.

$12 \text{ m/s}^2$

D.

$10 \text{ m/s}^2$

Physics-I Chapters

Kinematics-1D

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: